NCERT Solutions for Quadratic Equation Class 10 Maths Chapter 4 (2024)

Exercise 4.1:

This exercise covers the introduction to quadratic equations and the standard form of a quadratic equation. It also includes methods for solving quadratic equations by factorisation. In this exercise, students will learn how to identify a quadratic equation, how to convert a quadratic equation into standard form, and how to factorise quadratic equations using different methods. The exercise includes a set of questions that range from easy to difficult, allowing students to gradually build their understanding of the concepts.

Exercise 4.2:

This exercise covers more advanced methods for solving quadratic equations, such as completing the square and using the quadratic formula. It includes the derivation of the quadratic formula and shows how to apply it to solve quadratic equations. In this exercise, students will learn how to complete the square of a quadratic equation to convert it into standard form, and how to use the quadratic formula to solve quadratic equations. The exercise includes questions that require students to use both methods to solve quadratic equations.

Exercise 4.3:

This exercise focuses on the nature of the roots of quadratic equations and the discriminant of a quadratic equation. Students will learn how to determine the nature of the roots of a quadratic equation based on the value of its discriminant. The exercise covers the relationship between the coefficients and roots of a quadratic equation, and how to find the sum and product of the roots. The exercise includes a set of questions that require students to apply their knowledge of discriminant and the nature of roots to solve quadratic equations.

Exercise 4.4:

This exercise covers real-life applications of quadratic equations and includes word problems that require students to apply their knowledge of quadratic equations to solve practical problems. The exercise includes problems related to the trajectory of a projectile, finding the distance between two ships, and the dimensions of a garden. Students will learn how to formulate and solve quadratic equations to solve real-life problems. The exercise includes a set of word problems that gradually increase in difficulty, allowing students to develop their problem-solving skills.

Access NCERT Solutions for Class - 10 Maths Chapter 4 – Quadratic Equations

Exercise 4.1

1. Check whether the following are quadratic equations:

i. ${{\left( \text{x+1} \right)}^{\text{2}}}\text{=2}\left( \text{x-3} \right)$

Ans: ${{\left( \text{x+1} \right)}^{\text{2}}}\text{=2}\left( \text{x-3} \right)$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+2x+1=2x-6}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+7=0}$

Since, it is in the form of $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

Therefore, the given equation is a quadratic equation.

ii. ${{\text{x}}^{\text{2}}}\text{-2x=}\left( \text{-2} \right)\left( \text{3-x} \right)$

Ans: ${{\text{x}}^{\text{2}}}\text{-2x=}\left( \text{-2} \right)\left( \text{3-x} \right)$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-2x=-6+2x}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-4x+6=0}$

Since, it is in the form of $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

Therefore, the given equation is a quadratic equation.

iii. $\left( \text{x-2} \right)\left( \text{x+1} \right)\text{=}\left( \text{x-1} \right)\left( \text{x+3} \right)$

Ans: $\left( \text{x-2} \right)\left( \text{x+1} \right)\text{=}\left( \text{x-1} \right)\left( \text{x+3} \right)$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-x-2=}{{\text{x}}^{\text{2}}}\text{+2x-3}$

$\Rightarrow \text{3x-1=0}$

Since, it is not in the form of $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

Therefore, the given equation is not a quadratic equation.

iv. $\left( \text{x-3} \right)\left( \text{2x+1} \right)\text{=x}\left( \text{x+5} \right)$

Ans: $\left( \text{x-3} \right)\left( \text{2x+1} \right)\text{=x}\left( \text{x+5} \right)$

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{-5x-3=}{{\text{x}}^{\text{2}}}\text{+5x}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-10x-3=0}$

Since, it is in the form of $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

Therefore, the given equation is a quadratic equation.

v. $\left( \text{2x-1} \right)\left( \text{x-3} \right)\text{=}\left( \text{x+5} \right)\left( \text{x-1} \right)$

Ans: $\left( \text{2x-1} \right)\left( \text{x-3} \right)\text{=}\left( \text{x+5} \right)\left( \text{x-1} \right)$

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{-7x+3=}{{\text{x}}^{\text{2}}}\text{+4x-5}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-11x+8=0}$

Since, it is in the form of $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

Therefore, the given equation is a quadratic equation.

vi. ${{\text{x}}^{\text{2}}}\text{+3x+1=}{{\left( \text{x-2} \right)}^{\text{2}}}$

Ans: ${{\text{x}}^{\text{2}}}\text{+3x+1=}{{\left( \text{x-2} \right)}^{\text{2}}}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+3x+1=}{{\text{x}}^{\text{2}}}\text{+4-4x}$

$\Rightarrow \text{7x-3=0}$

Since, it is not in the form of $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

Therefore, the given equation is not a quadratic equation.

vii. ${{\left( \text{x+2} \right)}^{\text{3}}}\text{=2x}\left( {{\text{x}}^{\text{2}}}\text{-1} \right)$

Ans: ${{\left( \text{x+2} \right)}^{\text{3}}}\text{=2x}\left( {{\text{x}}^{\text{2}}}\text{-1} \right)$

$\Rightarrow {{\text{x}}^{\text{3}}}\text{+8+6}{{\text{x}}^{\text{2}}}\text{+12x=2}{{\text{x}}^{\text{3}}}\text{-2x}$

$\Rightarrow {{\text{x}}^{\text{3}}}\text{-14x-6}{{\text{x}}^{\text{2}}}\text{-8=0}$

Since, it is not in the form of $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

Therefore, the given equation is not a quadratic equation.

viii. ${{\text{x}}^{\text{3}}}\text{-4}{{\text{x}}^{\text{2}}}\text{-x+1=}{{\left( \text{x-2} \right)}^{\text{3}}}$

Ans: ${{\text{x}}^{\text{3}}}\text{-4}{{\text{x}}^{\text{2}}}\text{-x+1=}{{\left( \text{x-2} \right)}^{\text{3}}}$

$\Rightarrow {{\text{x}}^{\text{3}}}\text{-4}{{\text{x}}^{\text{2}}}\text{-x+1=}{{\text{x}}^{\text{3}}}\text{-8-6}{{\text{x}}^{\text{2}}}\text{+12x}$

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{-13x+9=0}$

Since, it is in the form of $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

Therefore, the given equation is a quadratic equation.

2. Represent the following situations in the form of quadratic equations.

i. The area of a rectangular plot is $\text{528 }{{\text{m}}^{\text{2}}}$. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

Ans: Let the breath of the plot be $\text{x m}$.

Thus, length would be-

$\text{Length=}\left( \text{2x+1} \right)\text{m}$

Hence, Area of rectangle $=$$\text{Length }\!\!\times\!\!\text{ breadth}$

So, $\text{528=x}\left( \text{2x+1} \right)$

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{+x-528=0}$

ii. The product of two consecutive positive integers is $\text{306}$. We need to find the integers.

Ans: Let the consecutive integers be $\text{x}$ and $\text{x+1}$.

Thus, according to question-

$\text{x}\left( \text{x+1} \right)\text{=306}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+x-306=0}$

iii. Rohan’s mother is $\text{26}$ years older than him. The product of their ages (in years) $\text{3}$ years from now will be $\text{360}$. We would like to find Rohan’s present age.

Ans: Let Rohan’s age be $\text{x}$.

Hence, his mother’s age is $\text{x+26}$ .

Now, after $\text{3 years}$.

Rohan’s age will be $\text{x+3}$.

His mother’s age will be $\text{x+29}$ .

So, according to question-

$\left( \text{x+3} \right)\left( \text{x+29} \right)\text{=360}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+3x+29x+87=360}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+32x-273=0}$

iv. A train travels a distance of $\text{480 km}$ at a uniform speed. If the speed had been $\text{8km/h}$ less, then it would have taken $\text{3}$ hours more to cover the same distance. We need to find the speed of the train.

Ans: Let the speed of train be $\text{x km/h}$.

Thus, time taken to travel $\text{482 km}$ is $\dfrac{\text{480}}{\text{x}}\text{hrs}$.

Now, let the speed of train $\text{=}\left( \text{x-8} \right)\text{km/h}$.

Therefore, time taken to travel $\text{480 km}$ is $\left( \dfrac{\text{480}}{\text{x}}+3 \right)\text{hrs}$.

Hence, $\text{speed }\!\!\times\!\!\text{ time=distance}$

i.e $\left( \text{x-8} \right)\left( \dfrac{\text{480}}{\text{x}}\text{+3} \right)\text{=480}$

$\Rightarrow \text{480+3x-}\dfrac{\text{3840}}{\text{x}}\text{-24=480}$

$\Rightarrow \text{3x-}\dfrac{\text{3840}}{\text{x}}\text{=24}$

$\Rightarrow \text{3}{{\text{x}}^{\text{2}}}\text{-24x-3840=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-8x-1280=0}$

Exercise 4.2

1. Find the roots of the following quadratic equations by factorisation:

i. ${{\text{x}}^{\text{2}}}\text{-3x-10=0}$

Ans: ${{\text{x}}^{\text{2}}}\text{-3x-10=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-5x+2x-10}$

$\Rightarrow \text{x}\left( \text{x-5} \right)\text{+2}\left( \text{x-5} \right)$

$\Rightarrow \left( \text{x-5} \right)\left( \text{x+2} \right)$

Therefore, roots of this equation are –

$\text{x-5=0}$ or $\text{x+2=0}$

i.e $\text{x=5}$ or $\text{x=-2}$

ii. $\text{2}{{\text{x}}^{\text{2}}}\text{+x-6=0}$

Ans: $\text{2}{{\text{x}}^{\text{2}}}\text{+x-6=0}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{+4x-3x-6}$

$\Rightarrow 2\text{x}\left( \text{x+2} \right)-3\left( \text{x+2} \right)$

$\Rightarrow \left( \text{x+2} \right)\left( \text{2x-3} \right)$

Therefore, roots of this equation are –

$\text{x+2=0}$ or $\text{2x-3=0}$

i.e $\text{x=-2}$ or $\text{x=}\dfrac{3}{2}$

iii. $\sqrt{\text{2}}{{\text{x}}^{\text{2}}}\text{+7x+5}\sqrt{\text{2}}\text{=0}$

Ans: $\sqrt{\text{2}}{{\text{x}}^{\text{2}}}\text{+7x+5}\sqrt{\text{2}}\text{=0}$

$\Rightarrow \sqrt{\text{2}}{{\text{x}}^{\text{2}}}\text{+5x+2x+5}\sqrt{\text{2}}$

$\Rightarrow \text{x}\left( \sqrt{\text{2}}\text{x+5} \right)+\sqrt{\text{2}}\left( \sqrt{\text{2}}\text{x+5} \right)$

$\Rightarrow \left( \sqrt{\text{2}}\text{x+5} \right)\left( \text{x+}\sqrt{\text{2}} \right)$

Therefore, roots of this equation are –

$\sqrt{\text{2}}\text{x+5=0}$ or $\text{x+}\sqrt{\text{2}}\text{=0}$

i.e $\text{x=}\dfrac{-5}{\sqrt{\text{2}}}$ or $\text{x=-}\sqrt{\text{2}}$

iv. $\text{2}{{\text{x}}^{\text{2}}}\text{-x+}\dfrac{\text{1}}{\text{8}}\text{=0}$

Ans: $\text{2}{{\text{x}}^{\text{2}}}\text{-x+}\dfrac{\text{1}}{\text{8}}\text{=0}$

\[\Rightarrow \dfrac{\text{1}}{\text{8}}\left( 16{{\text{x}}^{\text{2}}}-8x+1 \right)\]

\[\Rightarrow \dfrac{\text{1}}{\text{8}}\left( 4x\left( 4x-1 \right)-1\left( 4x-1 \right) \right)\]

$\Rightarrow \dfrac{\text{1}}{\text{8}} {{\left( \text{4x-1} \right)}^{2}}$

Therefore, roots of this equation are –

$\text{4x-1=0}$ or $\text{4x-1=0}$

i.e $\text{x=}\dfrac{1}{4}$ or $\text{x=}\dfrac{1}{4}$

v. $\text{100}{{\text{x}}^{\text{2}}}\text{-20x+1=0}$

Ans: $\text{100}{{\text{x}}^{\text{2}}}\text{-20x+1=0}$

$\Rightarrow 100{{\text{x}}^{\text{2}}}\text{-10x-10x+1}$

$\Rightarrow 10\text{x}\left( \text{10x-1} \right)-1\left( \text{10x-1} \right)$

\[\Rightarrow \left( \text{10x-1} \right)\left( \text{10x-1} \right)\]

Therefore, roots of this equation are –

\[\left( \text{10x-1} \right)=0\]or \[\left( \text{10x-1} \right)=0\]

i.e $\text{x=}\dfrac{1}{10}$ or $\text{x=}\dfrac{1}{10}$

2. Solve the problems given in Example 1

i. John and Jivanti together have $\text{45}$ marbles. Both of them lost $\text{5}$ marbles each, and the product of the number of marbles they now have is $\text{124}$. Find out how many marbles they had to start with.

Ans: Let the number of john’s marbles be $\text{x}$.

Thus, number of Jivanti’s marble be $\text{45-x}$.

According to question i.e,

After losing $\text{5}$ marbles.

Number of john’s marbles be $\text{x-5}$

And number of Jivanti’s marble be $\text{40-x}$.

Therefore, $\left( \text{x-5} \right)\left( \text{40-x} \right)\text{=124}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-45x+324=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-36x-9x+324=0}$

$\Rightarrow \text{x}\left( \text{x-36} \right)\text{-9}\left( \text{x-36} \right)\text{=0}$

$\Rightarrow \left( \text{x-36} \right)\left( \text{x-9} \right)\text{=0}$

So now,

Case 1- If $\text{x-36=0}$ i.e $\text{x=36}$

So, the number of john’s marbles be $\text{36}$.

Thus, number of Jivanti’s marble be $\text{9}$.

Case 2- If $\text{x-9=0}$ i.e $\text{x=9}$

So, the number of john’s marbles be $9$.

Thus, number of Jivanti’s marble be $36$.

ii. A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be $\text{55}$ minus the number of toys produced in a day. On a particular day, the total cost of production was Rs $\text{750}$. Find out the number of toys produced on that day.

Ans: Let the number of toys produced be $\text{x}$.

Therefore, Cost of production of each toy be $\text{Rs}\left( \text{55-x} \right)$.

Thus, $\left( \text{55-x} \right)\text{x=750}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-55x+750=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-25x-30x+750=0}$

$\Rightarrow \text{x}\left( \text{x-25} \right)-30\left( \text{x-25} \right)\text{=0}$

$\Rightarrow \left( \text{x-25} \right)\left( \text{x-30} \right)\text{=0}$

Case 1- If $\text{x-25=0}$ i.e $\text{x=25}$

So, the number of toys be $25$.

Case 2- If $\text{x-30=0}$ i.e $\text{x=30}$

So, the number of toys be $30$.

3. Find two numbers whose sum is $\text{27}$ and product is $\text{182}$.

Ans: Let the first number be $\text{x}$ ,

Thus, the second number be $\text{27-x}$.

Therefore,

$\text{x}\left( \text{27-x} \right)\text{=182}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-27x+182=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-13x-14x+182=0}$

$\Rightarrow \text{x}\left( \text{x-13} \right)-14\left( \text{x-13} \right)\text{=0}$

$\Rightarrow \left( \text{x-13} \right)\left( \text{x-14} \right)\text{=0}$

Case 1- If $\text{x-13=0}$ i.e $\text{x=13}$

So, the first number be $13$ ,

Thus, the second number be $\text{14}$.

Case 2- If $\text{x-14=0}$ i.e $\text{x=14}$

So, the first number be $\text{14}$.

Thus, the second number be$13$.

4. Find two consecutive positive integers, sum of whose squares is $\text{365}$.

Ans: Let the consecutive positive integers be $\text{x}$ and $\text{x+1}$.

Thus, ${{\text{x}}^{\text{2}}}\text{+}{{\left( \text{x+1} \right)}^{\text{2}}}\text{=365}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}{{\text{x}}^{\text{2}}}+1+2\text{x=365}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{+2x-364=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+x-182=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+14x-13x-182=0}$

$\Rightarrow \text{x}\left( \text{x+14} \right)-13\left( \text{x+14} \right)\text{=0}$

$\Rightarrow \left( \text{x+14} \right)\left( \text{x-13} \right)\text{=0}$

Case 1- If $\text{x+14=0}$ i.e $\text{x=-14}$.

This case is rejected because number is positive.

Case 2- If $\text{x-13=0}$ i.e $\text{x=13}$

So, the first number be $\text{13}$.

Thus, the second number be $14$.

Hence, the two consecutive positive integers are $\text{13}$ and $14$.

5. The altitude of a right triangle is $\text{7 cm}$ less than its base. If the hypotenuse is $\text{13 cm}$, find the other two sides.

Ans: Let the base of the right-angled triangle be $\text{x cm}$.

Its altitude be $\left( \text{x-7} \right)\text{cm}$.

Thus, by pythagores theorem-

$\text{bas}{{\text{e}}^{\text{2}}}\text{+altitud}{{\text{e}}^{\text{2}}}\text{=hypotenus}{{\text{e}}^{\text{2}}}$

\[\therefore {{\text{x}}^{\text{2}}}\text{+}{{\left( \text{x-7} \right)}^{\text{2}}}\text{=1}{{\text{3}}^{\text{2}}}\]

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}{{\text{x}}^{\text{2}}}+49-14\text{x=169}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{-14x-120=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-7x-60=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+12x+5x-60=0}$

$\Rightarrow \text{x}\left( \text{x-12} \right)+5\left( \text{x-12} \right)\text{=0}$

$\Rightarrow \left( \text{x-12} \right)\left( \text{x+5} \right)\text{=0}$

Case 1- If $\text{x-12=0}$ i.e $\text{x=12}$.

So, the base of the right-angled triangle be $\text{12 cm}$ and Its altitude be $\text{5cm}$

Case 2- If $\text{x+5=0}$ i.e $\text{x=-5}$

This case is rejected because side is always positive.

Hence, the base of the right-angled triangle be $\text{12 cm}$ and Its altitude be $\text{5cm}$.

6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was $\text{3}$ more than twice the number of articles produced on that day. If the total cost of production on that day was Rs $\text{90}$, find the number of articles produced and the cost of each article.

Ans: Let the number of articles produced be $\text{x}$.

Therefore, cost of production of each article be $\text{Rs}\left( \text{2x+3} \right)$.

Thus, $\text{x}\left( \text{2x+3} \right)\text{=90}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{+3x-90=0}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{+15x-12x-90=0}$

$\Rightarrow \text{x}\left( \text{2x+15} \right)-6\left( \text{2x+15} \right)\text{=0}$

$\Rightarrow \left( \text{2x+15} \right)\left( \text{x-6} \right)\text{=0}$

Case 1- If $\text{2x-15=0}$ i.e $\text{x=}\dfrac{-15}{2}$.

This case is rejected because number of articles is always positive.

Case 2- If $\text{x-6=0}$ i.e $\text{x=6}$

Hence, the number of articles produced be $6$.

Therefore, cost of production of each article be $\text{Rs15}$.

Exercise 4.3

1. Find the nature of the roots of the following quadratic equations.

If the real roots exist, find them-

i. $\text{2}{{\text{x}}^{\text{2}}}\text{-3x+5=0}$

Ans: For a quadratic equation $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

Where Discriminant $\text{=}{{\text{b}}^{\text{2}}}\text{-4ac}$

Then –

Case 1- If ${{\text{b}}^{\text{2}}}\text{-4ac>0}$ then there will be two distinct real roots.

Case 2- If ${{\text{b}}^{\text{2}}}\text{-4ac=0}$ then there will be two equal real roots.

Case 3- If ${{\text{b}}^{\text{2}}}\text{-4ac<0}$ then there will be no real roots.

Thus, for $\text{2}{{\text{x}}^{\text{2}}}\text{-3x+5=0}$ .

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=2}$, $\text{b=-3}$, $\text{c=5}$.

Discriminant $\text{=}{{\left( \text{-3} \right)}^{\text{2}}}\text{-4}\left( \text{2} \right)\left( \text{5} \right)$

$\text{=9-40}$

$\text{=-31}$

Since, Discriminant: ${{\text{b}}^{\text{2}}}\text{-4ac < 0}$.

Therefore, there is no real root for the given equation.

ii. $\text{3}{{\text{x}}^{\text{2}}}\text{-4}\sqrt{\text{3}}\text{x+4=0}$

Ans: For a quadratic equation $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

Where Discriminant $\text{=}{{\text{b}}^{\text{2}}}\text{-4ac}$

Then –

Case 1- If ${{\text{b}}^{\text{2}}}\text{-4ac > 0}$ then there will be two distinct real roots.

Case 2- If ${{\text{b}}^{\text{2}}}\text{-4ac=0}$ then there will be two equal real roots.

Case 3- If ${{\text{b}}^{\text{2}}}\text{-4ac < 0}$ then there will be no real roots.

Thus, for $\text{3}{{\text{x}}^{\text{2}}}\text{-4}\sqrt{\text{3}}\text{x+4=0}$ .

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=3}$, $\text{b=-4}\sqrt{\text{3}}$, $\text{c=4}$.

Discriminant $\text{=}{{\left( \text{-4}\sqrt{\text{3}} \right)}^{\text{2}}}\text{-4}\left( \text{3} \right)\left( \text{4} \right)$

$\text{=48-48}$

$\text{=0}$

Since, Discriminant: ${{\text{b}}^{\text{2}}}\text{-4ac=0}$.

Therefore, there is equal real root for the given equation and the roots are-

$\dfrac{\text{-b}}{\text{2a}}$ and $\dfrac{\text{-b}}{\text{2a}}$.

Hence, roots are-

$\dfrac{\text{-b}}{\text{2a}}\text{=}\dfrac{\text{-}\left( \text{-4}\sqrt{\text{3}} \right)}{\text{6}}$

$\text{=}\dfrac{\text{4}\sqrt{\text{3}}}{\text{6}}$

\[\text{=}\dfrac{\text{2}\sqrt{\text{3}}}{3}\]

Therefore, roots are \[\dfrac{\text{2}\sqrt{\text{3}}}{3}\] and \[\dfrac{\text{2}\sqrt{\text{3}}}{3}\].

iii. $\text{2}{{\text{x}}^{\text{2}}}\text{-6x+3=0}$

Ans: For a quadratic equation $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

Where Discriminant $\text{=}{{\text{b}}^{\text{2}}}\text{-4ac}$

Then –

Case 1- If ${{\text{b}}^{\text{2}}}\text{-4ac>0}$ then there will be two distinct real roots.

Case 2- If ${{\text{b}}^{\text{2}}}\text{-4ac=0}$ then there will be two equal real roots.

Case 3- If ${{\text{b}}^{\text{2}}}\text{-4ac<0}$ then there will be no real roots.

Thus, for $\text{2}{{\text{x}}^{\text{2}}}\text{-6x+3=0}$ .

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=2}$, $\text{b=-6}$, $\text{c=3}$.

Discriminant $\text{=}{{\left( \text{-6} \right)}^{\text{2}}}\text{-4}\left( \text{2} \right)\left( \text{3} \right)$

$\text{=36-24}$

$\text{=12}$

Since, Discriminant: ${{\text{b}}^{\text{2}}}\text{-4ac>0}$.

Therefore, distinct real roots exists for the given equation and the roots are-

$\text{x=}\dfrac{\text{-b }\!\!\pm\!\!\text{ }\sqrt{{{\text{b}}^{\text{2}}}\text{-4ac}}}{\text{2a}}$

Hence, roots are-

$\text{x=}\dfrac{\text{-}\left( \text{-6} \right)\text{ }\!\!\pm\!\!\text{ }\sqrt{{{\left( \text{-6} \right)}^{\text{2}}}\text{-4}\left( \text{2} \right)\left( \text{3} \right)}}{\text{4}}$

$\text{=}\dfrac{\text{6 }\!\!\pm\!\!\text{ }\sqrt{\text{36-24}}}{\text{4}}$

$\text{=}\dfrac{\text{6 }\!\!\pm\!\!\text{ }\sqrt{\text{12}}}{\text{4}}$

$\text{=}\dfrac{\text{6 }\!\!\pm\!\!\text{ 2}\sqrt{\text{3}}}{\text{4}}$

$\text{=}\dfrac{\text{3 }\!\!\pm\!\!\text{ }\sqrt{\text{3}}}{\text{2}}$

Therefore, roots are $\dfrac{\text{3+}\sqrt{\text{3}}}{\text{2}}$ and $\dfrac{\text{3-}\sqrt{\text{3}}}{\text{2}}$.

2. Find the values of $\text{k}$ for each of the following quadratic equations, so

that they have two equal roots.

i. $\text{2}{{\text{x}}^{\text{2}}}\text{+kx+3=0}$

Ans: If a quadratic equation $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$ has two equal roots, then its discriminant will be $\text{0}$ i.e., ${{\text{b}}^{\text{2}}}\text{-4ac=0}$

So, for $\text{2}{{\text{x}}^{\text{2}}}\text{+kx+3=0}$.

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=2}$, $\text{b=k}$, $\text{c=3}$.

Discriminant $\text{=}{{\left( \text{k} \right)}^{\text{2}}}\text{-4}\left( \text{2} \right)\left( \text{3} \right)$

$\text{=}{{\text{k}}^{2}}-24$

For equal roots-

${{\text{b}}^{\text{2}}}\text{-4ac=0}$

$\therefore {{\text{k}}^{\text{2}}}\text{-24=0}$

$\Rightarrow {{\text{k}}^{\text{2}}}\text{=24}$

$\Rightarrow \text{k=}\sqrt{\text{24}}$

$\Rightarrow \text{k=}\pm \text{2}\sqrt{\text{6}}$

ii. $\text{kx}\left( \text{x-2} \right)\text{+6=0}$

Ans: If a quadratic equation $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$ has two equal roots, then its discriminant will be $\text{0}$ i.e., ${{\text{b}}^{\text{2}}}\text{-4ac=0}$

So, for $\text{kx}\left( \text{x-2} \right)\text{+6=0}$

$\Rightarrow \text{k}{{\text{x}}^{\text{2}}}\text{-2kx+6=0}$

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=k}$, $\text{b=-2k}$, $\text{c=6}$.

Discriminant $\text{=}{{\left( \text{-2k} \right)}^{\text{2}}}\text{-4}\left( \text{k} \right)\left( \text{6} \right)$

$\text{=4}{{\text{k}}^{\text{2}}}\text{-24k}$

For equal roots-

${{\text{b}}^{\text{2}}}\text{-4ac=0}$

$\therefore \text{4}{{\text{k}}^{\text{2}}}\text{-24k=0}$

$\Rightarrow \text{4k}\left( \text{k-6} \right)\text{=0}$

$\Rightarrow \text{k=0 or k=6}$

But $\text{k}$ cannot be zero. Thus, this equation has two equal roots when $\text{k}$ should be $\text{6}$ .

3. Is it possible to design a rectangular mango grove whose length is

twice its breadth, and the area is $\text{800}{{\text{m}}^{\text{2}}}$ ? If so, find its length and breadth.

Ans: Let the breadth of mango grove be $\text{x}$.

So, length of mango grove will be $\text{2x}$.

Hence, Area of mango grove is $=\left( \text{2x} \right)\text{x}$

$\text{=2}{{\text{x}}^{\text{2}}}$.

So, $\text{2}{{\text{x}}^{\text{2}}}\text{=800}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{=400}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-400=0}$

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=1}$, $\text{b=0}$, $\text{c=400}$.

Discriminant $\text{=}{{\left( \text{0} \right)}^{\text{2}}}\text{-4}\left( \text{1} \right)\left( \text{-400} \right)$

$\text{=1600}$

Since, Discriminant: ${{\text{b}}^{\text{2}}}\text{-4ac>0}$.

Therefore, distinct real roots exist for the given equation and the roots are-

$\text{x=}\dfrac{\text{-b }\!\!\pm\!\!\text{ }\sqrt{{{\text{b}}^{\text{2}}}\text{-4ac}}}{\text{2a}}$

Hence, roots are-

$\text{x=}\dfrac{\text{-}\left( 0 \right)\text{ }\!\!\pm\!\!\text{ }\sqrt{{{\left( 0 \right)}^{\text{2}}}\text{-4}\left( 1 \right)\left( -400 \right)}}{2}$

$\text{=}\dfrac{\pm \sqrt{\text{1600}}}{2}$

$\text{=}\dfrac{\text{ }\!\!\pm\!\!\text{ 40}}{2}$

$\text{=}\pm \text{20}$

Since, length cannot be negative.

Therefore, breadth of the mango grove is $\text{20m}$.

And length of the mango grove be $\text{2}\left( \text{20} \right)\text{m}$ i.e., $\text{40m}$.

4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is $\text{20}$ years. Four years ago, the product of their ages in years was $\text{48}$.

Ans: Let the age of one friend be $\text{x years}$.

So, age of the other friend will be $\left( \text{20-x} \right)\text{years}$.

Thus, four years ago, the age of one friend be $\left( \text{x-4} \right)\text{years}$.

And age of the other friend will be $\left( \text{16-x} \right)\text{years}$.

Hence, according to question-

$\left( \text{x-4} \right)\left( \text{16-x} \right)\text{=48}$

$\Rightarrow \text{16x-64-}{{\text{x}}^{\text{2}}}\text{+4x=48}$

$\Rightarrow 20\text{x-112-}{{\text{x}}^{\text{2}}}\text{=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-20x+112-=0}$

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=1}$, $\text{b=-20}$, $\text{c=112}$.

Discriminant $\text{=}{{\left( \text{-20} \right)}^{\text{2}}}\text{-4}\left( \text{1} \right)\left( \text{112} \right)$

$\text{=400-448}$

$\text{=-48}$

Since, Discriminant: ${{\text{b}}^{\text{2}}}\text{-4ac <0}$.

Therefore, there is no real root for the given equation and hence, this situation is not possible.

5. Is it possible to design a rectangular park of perimeter $\text{80 m}$ and area $\text{400}{{\text{m}}^{\text{2}}}$? If so find its length and breadth.

Ans: Let the length of the park be $\text{x m}$ and breadth of the park be $\text{x m}$.

Thus, $\text{Perimeter=2}\left( \text{x+y} \right)$.

Hence, according to question-

$\text{2}\left( \text{x+y} \right)\text{=80}$

$\Rightarrow \text{x+y=40}$

$\Rightarrow \text{y=40-x}$.

Now, $\text{Area=x }\!\!\times\!\!\text{ y}$.

Substituting value of y.

$\text{Area=x}\left( \text{40-x} \right)$

So, according to question-

$\text{x}\left( \text{40-x} \right)\text{=400}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-40x+400=0}$

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=1}$, $\text{b=-40}$, $\text{c=400}$.

Discriminant $\text{=}{{\left( \text{-40} \right)}^{\text{2}}}\text{-4}\left( \text{1} \right)\left( 400 \right)$

$\text{=1600-1600}$

$\text{=0}$

Since, Discriminant: ${{\text{b}}^{\text{2}}}\text{-4ac=0}$.

Therefore, there is equal real roots for the given equation and hence, this situation is possible.

Hence, roots are-

$\dfrac{\text{-b}}{\text{2a}}\text{=}\dfrac{\text{-}\left( -40 \right)}{2}$

$\text{=}\dfrac{\text{40}}{2}$

\[\text{=20}\]

Therefore, length of park is $\text{x=20m}$ .

And breadth of park be $\text{y=}\left( \text{40-20} \right)\text{m}$ i.e., $\text{y=20m}$.

Exercise (Not in the current syllabus)

1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

i. $\text{2}{{\text{x}}^{\text{2}}}\text{-7x+3=0}$

Ans: $\text{2}{{\text{x}}^{\text{2}}}\text{-7x+3=0}$

On dividing both sides of the equation by $\text{2}$.

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-}\dfrac{\text{7}}{\text{2}}\text{x=-}\dfrac{\text{3}}{\text{2}}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-2}\left( \dfrac{\text{7}}{\text{4}} \right)\text{x=-}\dfrac{\text{3}}{\text{2}}$

On adding ${{\left( \dfrac{7}{4} \right)}^{2}}$ both sides of equation.

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-2}\left( \dfrac{\text{7}}{\text{4}} \right)\text{x+}{{\left( \dfrac{7}{4} \right)}^{2}}\text{=-}\dfrac{\text{3}}{\text{2}}+{{\left( \dfrac{7}{4} \right)}^{2}}$

$\Rightarrow {{\left( \text{x-}\dfrac{\text{7}}{\text{4}} \right)}^{\text{2}}}\text{=}\dfrac{\text{49}}{\text{16}}\text{-}\dfrac{\text{3}}{\text{2}}$

$\Rightarrow {{\left( \text{x-}\dfrac{\text{7}}{\text{4}} \right)}^{\text{2}}}\text{=}\dfrac{\text{25}}{\text{16}}$

$\Rightarrow \left( \text{x-}\dfrac{\text{7}}{\text{4}} \right)\text{=}\pm \dfrac{5}{4}$

$\Rightarrow x\text{=}\dfrac{\text{7}}{\text{4}}\pm \dfrac{5}{4}$

$\Rightarrow x\text{=}\dfrac{\text{7}}{\text{4}}+\dfrac{5}{4}$ or $x\text{=}\dfrac{\text{7}}{\text{4}}-\dfrac{5}{4}$

$\Rightarrow x\text{=}\dfrac{12}{\text{4}}$ or $x\text{=}\dfrac{2}{\text{4}}$

$\Rightarrow x\text{=}3$ or $x\text{=}\dfrac{\text{1}}{\text{2}}$

ii. $\text{2}{{\text{x}}^{\text{2}}}\text{+x-4=0}$

Ans: $\text{2}{{\text{x}}^{\text{2}}}\text{+x-4=0}$

On dividing both sides of the equation by $\text{2}$.

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}\dfrac{1}{2}\text{x=2}$

On adding ${{\left( \dfrac{1}{4} \right)}^{2}}$ both sides of equation.

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+2}\left( \dfrac{1}{4} \right)\text{x+}{{\left( \dfrac{1}{4} \right)}^{2}}\text{=2+}{{\left( \dfrac{1}{4} \right)}^{2}}$

$\Rightarrow {{\left( \text{x+}\dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{=}\dfrac{\text{33}}{\text{16}}$

\[\Rightarrow \left( \text{x+}\dfrac{\text{1}}{\text{4}} \right)\text{=}\pm \dfrac{\sqrt{\text{33}}}{4}\]

\[\Rightarrow \text{x= }\!\!\pm\!\!\text{ }\dfrac{\sqrt{\text{33}}}{\text{4}}\text{-}\dfrac{\text{1}}{\text{4}}\]

\[\Rightarrow \text{x= }\!\!\pm\!\!\text{ }\dfrac{\sqrt{\text{33}}-1}{\text{4}}\]

\[\Rightarrow \text{x=}\dfrac{\sqrt{\text{33}}-1}{\text{4}}\] or \[\Rightarrow \text{x=}\dfrac{\sqrt{\text{-33}}-1}{\text{4}}\]

iii. $\text{4}{{\text{x}}^{\text{2}}}\text{+4}\sqrt{3}\text{x+3=0}$

Ans: $\text{4}{{\text{x}}^{\text{2}}}\text{+4}\sqrt{3}\text{x+3=0}$

$\Rightarrow {{\left( 2\text{x} \right)}^{\text{2}}}\text{+2}\left( 2\sqrt{3} \right)\text{x+}{{\left( \sqrt{\text{3}} \right)}^{2}}\text{=0}$

$\Rightarrow {{\left( \text{2x+}\sqrt{\text{3}} \right)}^{\text{2}}}\text{=0}$

$\Rightarrow \left( \text{2x+}\sqrt{\text{3}} \right)\text{=0}$ and $\Rightarrow \left( \text{2x+}\sqrt{\text{3}} \right)\text{=0}$

$\Rightarrow \text{x=}\dfrac{\text{-}\sqrt{\text{3}}}{\text{2}}$ and $\Rightarrow \text{x=}\dfrac{\text{-}\sqrt{\text{3}}}{\text{2}}$

iv. $\text{2}{{\text{x}}^{\text{2}}}\text{+x+4=0}$

Ans: $\text{2}{{\text{x}}^{\text{2}}}\text{+x+4=0}$

On dividing both sides of the equation by $\text{2}$.

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}\dfrac{\text{1}}{\text{2}}\text{x+2=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+2}\left( \dfrac{\text{1}}{4} \right)\text{x=-2}$

On adding ${{\left( \dfrac{1}{4} \right)}^{2}}$ both sides of equation.

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+2}\left( \dfrac{\text{1}}{4} \right)\text{x+}{{\left( \dfrac{1}{4} \right)}^{2}}\text{=-2+}{{\left( \dfrac{1}{4} \right)}^{2}}$

$\Rightarrow {{\left( \text{x+}\dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{=}\dfrac{\text{1}}{\text{16}}\text{-2}$

$\Rightarrow {{\left( \text{x+}\dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{=-}\dfrac{\text{31}}{\text{16}}$

Since, the square of a number cannot be negative.

Therefore, there is no real root for the given equation.

2. Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.

i. $\text{2}{{\text{x}}^{\text{2}}}\text{-7x+3=0}$

Ans: $\text{2}{{\text{x}}^{\text{2}}}\text{-7x+3=0}$

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=2}$, $\text{b=-7}$, $\text{c=3}$.

Therefore, by using quadratic formula-

$\text{x=}\dfrac{\text{-b }\!\!\pm\!\!\text{ }\sqrt{{{\text{b}}^{\text{2}}}\text{-4ac}}}{\text{2a}}$

$\Rightarrow \text{x=}\dfrac{\text{7 }\!\!\pm\!\!\text{ }\sqrt{\text{49-24}}}{4}$

$\Rightarrow \text{x=}\dfrac{\text{7 }\!\!\pm\!\!\text{ 5}}{4}$

$\Rightarrow \text{x=}\dfrac{\text{7+5}}{4}$ or $\Rightarrow \text{x=}\dfrac{\text{7-5}}{4}$

$\Rightarrow \text{x=}\dfrac{12}{4}$ or $\Rightarrow \text{x=}\dfrac{\text{2}}{4}$

$\therefore \text{x=3 or }\dfrac{\text{1}}{\text{2}}$.

ii. $\text{2}{{\text{x}}^{\text{2}}}\text{+x-4=0}$

Ans: $\text{2}{{\text{x}}^{\text{2}}}\text{+x-4=0}$

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=2}$, $\text{b=1}$, $\text{c=-4}$.

Therefore, by using quadratic formula-

$\text{x=}\dfrac{\text{-b }\!\!\pm\!\!\text{ }\sqrt{{{\text{b}}^{\text{2}}}\text{-4ac}}}{\text{2a}}$

$\Rightarrow \text{x=}\dfrac{\text{-1 }\!\!\pm\!\!\text{ }\sqrt{\text{1-32}}}{4}$

$\Rightarrow \text{x=}\dfrac{\text{-1 }\!\!\pm\!\!\text{ }\sqrt{33}}{4}$

$\Rightarrow \text{x=}\dfrac{\text{-1+}\sqrt{33}}{4}$ or $\Rightarrow \text{x=}\dfrac{\text{-1-}\sqrt{33}}{4}$

$\therefore \text{x=}\dfrac{\text{-1+}\sqrt{33}}{4}\text{ or }\dfrac{\text{-1-}\sqrt{33}}{4}$.

iii. $\text{4}{{\text{x}}^{\text{2}}}\text{+4}\sqrt{3}\text{x+3=0}$

Ans: $\text{4}{{\text{x}}^{\text{2}}}\text{+4}\sqrt{3}\text{x+3=0}$

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=4}$, $\text{b=4}\sqrt{3}$, $\text{c=3}$.

Therefore, by using quadratic formula-

$\text{x=}\dfrac{\text{-b }\!\!\pm\!\!\text{ }\sqrt{{{\text{b}}^{\text{2}}}\text{-4ac}}}{\text{2a}}$

$\Rightarrow \text{x=}\dfrac{\text{-4}\sqrt{3}\text{ }\!\!\pm\!\!\text{ }\sqrt{\text{48-48}}}{8}$

$\Rightarrow \text{x=}\dfrac{\text{-4}\sqrt{3}}{8}$

$\Rightarrow \text{x=}\dfrac{\text{-}\sqrt{3}}{2}$ or $\Rightarrow \text{x=}\dfrac{\text{-}\sqrt{3}}{2}$

$\therefore \text{x=}\dfrac{\text{-}\sqrt{3}}{2}\text{ or }\dfrac{\text{-}\sqrt{3}}{2}$.

iv. $\text{2}{{\text{x}}^{\text{2}}}\text{+x+4=0}$

Ans: $\text{2}{{\text{x}}^{\text{2}}}\text{+x+4=0}$

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=2}$, $\text{b=1}$, $\text{c=4}$.

Therefore, by using quadratic formula-

$\text{x=}\dfrac{\text{-b }\!\!\pm\!\!\text{ }\sqrt{{{\text{b}}^{\text{2}}}\text{-4ac}}}{\text{2a}}$

\[\Rightarrow \text{x=}\dfrac{\text{-1 }\!\!\pm\!\!\text{ }\sqrt{\text{1-32}}}{4}\]

$\Rightarrow \text{x=}\dfrac{\text{-1 }\!\!\pm\!\!\text{ }\sqrt{\text{-31}}}{4}$

Since, there can not be any negative number inside square root for any real root to exist.

Therefore, there is no real root for the given equation.

3. Find the roots of the following equations:

i. $\text{x-}\dfrac{\text{1}}{\text{x}}\text{=3,x}\ne \text{0}$

Ans: $\text{x-}\dfrac{\text{1}}{\text{x}}\text{=3,x}\ne \text{0}$

$\Rightarrow\text{1}{{\text{x}}^{\text{2}}}\text{-3x-1=}0$.

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=1}$, $\text{b=-3}$, $\text{c=-1}$.

Therefore, by using quadratic formula-

$\text{x=}\dfrac{\text{-b }\!\!\pm\!\!\text{ }\sqrt{{{\text{b}}^{\text{2}}}\text{-4ac}}}{\text{2a}}$

$\Rightarrow \text{x=}\dfrac{\text{3 }\!\!\pm\!\!\text{ }\sqrt{\text{9+4}}}{2}$

$\Rightarrow \text{x=}\dfrac{\text{3 }\!\!\pm\!\!\text{ }\sqrt{\text{13}}}{2}$

$\Rightarrow \text{x=}\dfrac{\text{3+}\sqrt{\text{13}}}{2}$ or $\Rightarrow \text{x=}\dfrac{\text{3-}\sqrt{\text{13}}}{2}$

$\therefore \text{x=}\dfrac{\text{3+}\sqrt{\text{13}}}{2}\text{ or }\dfrac{\text{3-}\sqrt{\text{13}}}{2}$.

ii. $\dfrac{\text{1}}{\text{x+4}}\text{-}\dfrac{\text{1}}{\text{x-7}}\text{=}\dfrac{\text{11}}{\text{30}}\text{,x}\ne \text{-4,7}$

Ans: $\dfrac{\text{1}}{\text{x+4}}\text{-}\dfrac{\text{1}}{\text{x-7}}\text{=}\dfrac{\text{11}}{\text{30}}\text{,x}\ne \text{-4,7}$

$\Rightarrow \dfrac{\text{x-7-x-4}}{\left( \text{x+4} \right)\left( \text{x-7} \right)}\text{=}\dfrac{\text{11}}{\text{30}}$

$\Rightarrow \dfrac{\text{-11}}{\left( \text{x+4} \right)\left( \text{x-7} \right)}\text{=}\dfrac{\text{11}}{\text{30}}$

$\Rightarrow \left( \text{x+4} \right)\left( \text{x-7} \right)=-30$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-3x-28=-3}0$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-3x+2=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-2x-x+2=0}$

$\Rightarrow \text{x}\left( \text{x-2} \right)\text{-1}\left( \text{x-2} \right)\text{=0}$

$\Rightarrow \left( \text{x-2} \right)\left( \text{x-1} \right)\text{=0}$

$\Rightarrow x\text{=1 or 2}$

4. The sum of the reciprocals of Rehman’s ages, (in years) $\text{3}$ years ago

and $\text{5}$ years from now is \[\dfrac{\text{1}}{\text{3}}\]. Find his present age.

Ans: Let the present age of Rehman be $\text{x}$ years.

Three years ago, his age was $\left( \text{x-3} \right)\text{years}$.

Five years hence, his age will be $\left( \text{x+5} \right)\text{years}$.

Therefore,$\dfrac{\text{1}}{\text{x-3}}\text{+}\dfrac{\text{1}}{\text{x+5}}\text{=}\dfrac{\text{1}}{\text{3}}$

\[\Rightarrow \dfrac{\text{x+5+x-3}}{\left( \text{x-3} \right)\left( \text{x+5} \right)}\text{=}\dfrac{\text{1}}{\text{3}}\]

\[\Rightarrow \dfrac{\text{2x+2}}{\left( \text{x-3} \right)\left( \text{x+5} \right)}\text{=}\dfrac{\text{1}}{\text{3}}\]

$\Rightarrow 3\left( \text{2x+2} \right)=\left( \text{x-3} \right)\left( \text{x+5} \right)$

$\Rightarrow \text{6x+6=}{{\text{x}}^{\text{2}}}\text{+2x-15}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-4x-21=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-7x+3x-21=0}$

$\Rightarrow \text{x}\left( \text{x-7} \right)\text{+3}\left( \text{x-7} \right)\text{=0}$

$\Rightarrow \left( \text{x-7} \right)\left( \text{x+3} \right)\text{=0}$

$\Rightarrow x\text{=7 or -3}$

Therefore, Rehman’s age is $\text{7 years}$.

5. In a class test, the sum of Shefali’s marks in Mathematics and English is $\text{30}$. Had she got $\text{2}$ marks more in Mathematics and $\text{3}$ marks less in English, the product of their marks would have been $\text{210}$. Find her marks in the two subjects.

Ans: Let the marks in maths be $\text{x}$.

Thus, marks in English will be $\text{30-x}$.

Hence, according to question –

$\left( \text{x+2} \right)\left( \text{30-x-3} \right)\text{=210}$

$\left( \text{x+2} \right)\left( \text{27-x} \right)\text{=210}$

$\Rightarrow \text{-}{{\text{x}}^{\text{2}}}\text{+25x+54=210}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-25x+156=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-12x-13x+156=0}$

$\Rightarrow \text{x}\left( \text{x-12} \right)\text{-13}\left( \text{x-12} \right)\text{=0}$

$\Rightarrow \left( \text{x-12} \right)\left( \text{x-13} \right)\text{=0}$

$\Rightarrow \text{x=12,13}$

Case 1- If the marks in mathematics are $\text{12}$ , then marks in English will be $18$.

Case 2- If the marks in mathematics are $\text{13}$ , then marks in English will be $17$.

6. The diagonal of a rectangular field is $\text{60}$ metres more than the shorter side. If the longer side is $\text{30}$ metres more than the shorter side, find the sides of the field.

Ans: Let the shorter side of the rectangle be $\text{x m}$.

Thus, Larger side of the rectangle will be $\left( \text{x+30} \right)\text{m}$.

Diagonal of the rectangle be $\sqrt{{{\text{x}}^{\text{2}}}\text{+}{{\left( \text{x+30} \right)}^{\text{2}}}}$

Hence, according to question-

$\sqrt{{{\text{x}}^{\text{2}}}\text{+}{{\left( \text{x+30} \right)}^{\text{2}}}}\text{=x+60}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}{{\left( \text{x+30} \right)}^{\text{2}}}\text{=}{{\left( \text{x+60} \right)}^{2}}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}{{\text{x}}^{\text{2}}}\text{+900+60x=}{{\text{x}}^{\text{2}}}\text{+3600+120x}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-60x-2700=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-90x+30x-2700=0}$

$\Rightarrow \text{x}\left( \text{x-90} \right)+30\left( \text{x-90} \right)\text{=0}$

$\Rightarrow \left( \text{x-90} \right)\left( \text{x+30} \right)\text{=0}$

$\Rightarrow \text{x=90,-30}$

Since, side cannot be negative.

Therefore, the length of the shorter side of rectangle is $\text{90 m}$.

Hence, length of the larger side of the rectangle be $\text{120 m}$.

7. The difference of squares of two numbers is $\text{180}$. The square of the smaller number is $\text{8}$ times the larger number. Find the two numbers.

Ans: Let the larger number be $\text{x}$ and smaller number be $\text{y}$.

According to question-

${{\text{x}}^{\text{2}}}\text{-}{{\text{y}}^{\text{2}}}\text{=180}$ and ${{\text{y}}^{\text{2}}}\text{=8x}$

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{-8x=180}\]

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{-8x-180=0}\]

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{-18x+10x-180=0}\]

$\Rightarrow \text{x}\left( \text{x-18} \right)+10\left( \text{x-18} \right)\text{=0}$

$\Rightarrow \left( \text{x-18} \right)\left( \text{x+10} \right)\text{=0}$

$\Rightarrow \text{x=18,-10}$

Since, larger cannot be negative as $8$ times of the larger number will be negative and hence, the square of the smaller number will be negative which is not possible.

Therefore, the larger number will be $18$.

$\therefore {{\text{y}}^{\text{2}}}\text{=8}\left( \text{18} \right)$

$\Rightarrow {{\text{y}}^{\text{2}}}\text{=144}$

$\Rightarrow \text{y= }\!\!\pm\!\!\text{ 12}$

Hence, smaller number be $\pm 12$.

Therefore, the numbers are $18$ and $12$ or $18$ and $-12$ .

8. A train travels $\text{360 km}$km at a uniform speed. If the speed had been $\text{5km/h}$ more, it would have taken $\text{1}$hour less for the same journey. Find the speed of the train.

Ans: Let the speed of the train be $\text{x km/h}$.

Time taken to cover $\text{360 km/h}$ be $\dfrac{\text{360}}{\text{x}}$.

According to question-

$\left( \text{x+5} \right)\left( \dfrac{\text{360}}{\text{x}}\text{-1} \right)\text{=360}$

$\Rightarrow \text{360-x+}\dfrac{\text{1800}}{\text{x}}\text{-5=360}$

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{+5x-1800=0}\]

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{+45x-40x-1800=0}\]

$\Rightarrow \text{x}\left( \text{x+45} \right)-40\left( \text{x+45} \right)\text{=0}$

$\Rightarrow \left( \text{x+45} \right)\left( \text{x-40} \right)\text{=0}$

$\Rightarrow \text{x=40,-45}$

Since, the speed cannot be negative.

Therefore, the speed of the train is $\text{40 km/h}$.

9. Two water taps together can fill a tank in $\text{9}\dfrac{\text{3}}{\text{8}}$ hours. The tap of larger diameter takes $\text{10}$ hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Ans: Let the time taken by the smaller pipe to fill the tank be $\text{x hr}$.

So, time taken by larger pipe be $\left( \text{x-10} \right)\text{hr}$.

Part of the tank filled by smaller pipe in $1$ hour is $\dfrac{\text{1}}{\text{x}}$.

Part of the tank filled by larger pipe in $1$ hour is $\dfrac{\text{1}}{\text{x-10}}$.

So, according to the question-

$\dfrac{\text{1}}{\text{x}}\text{+}\dfrac{\text{1}}{\text{x-10}}\text{=9}\dfrac{\text{3}}{\text{8}}$

$\Rightarrow \dfrac{\text{1}}{\text{x}}\text{+}\dfrac{\text{1}}{\text{x-10}}\text{=}\dfrac{\text{75}}{\text{8}}$

\[\Rightarrow \dfrac{\text{x-10+x}}{\text{x}\left( \text{x-10} \right)}\text{=}\dfrac{\text{8}}{\text{75}}\]

\[\Rightarrow \dfrac{\text{2x-10}}{\text{x}\left( \text{x-10} \right)}\text{=}\dfrac{\text{8}}{\text{75}}\]

$\Rightarrow \text{75}\left( \text{2x-10} \right)\text{=8}{{\text{x}}^{\text{2}}}\text{-80x}$

$\Rightarrow \text{150x-750=8}{{\text{x}}^{\text{2}}}\text{-80x}$

$\Rightarrow \text{8}{{\text{x}}^{\text{2}}}\text{-230x+750=0}$

$\Rightarrow \text{8}{{\text{x}}^{\text{2}}}\text{-200x-30x+750=0}$

$\Rightarrow \text{8x}\left( \text{x-25} \right)\text{-30}\left( \text{x-25} \right)\text{=0}$

$\Rightarrow \left( \text{x-25} \right)\left( \text{8x-30} \right)\text{=0}$

$\Rightarrow x\text{=25 or }\dfrac{\text{30}}{\text{8}}$

Case 1- If time taken by smaller pipe be $\dfrac{\text{30}}{\text{8}}$ i.e $\text{3}\text{.75 hours}$.

So, Time taken by larger pipe will be negative which is not possible.

Hence, this case is rejected.

Case 2- If the time taken by smaller pipe be $\text{25}$.Then, time taken by larger pipe will be $\text{15 hours}$.

Therefore, time taken by smaller pipe be $\text{25 hours}$ and time taken by larger pipe will be $\text{15 hours}$.

10. An express train takes $\text{1}$ hour less than a passenger train to travel $\text{132 km}$ between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speeds of the express train is $\text{11 km/h}$ more than that of the passenger train, find the average speed of the two trains.

Ans: Let the average speed of passenger train be $\text{x km/h}$.

So, Average speed of express train be $\left( \text{x+11} \right)\text{km/h}$.

Thus, according to question.

$\therefore \dfrac{\text{132}}{\text{x}}\text{-}\dfrac{\text{132}}{\text{x+11}}\text{=1}$

$\Rightarrow \text{132}\left[ \dfrac{\text{x+11-x}}{\text{x}\left( \text{x+11} \right)} \right]\text{=1}$

$\Rightarrow \dfrac{\text{132 }\!\!\times\!\!\text{ 11}}{\text{x}\left( \text{x+11} \right)}\text{=1}$

$\Rightarrow \text{132 }\!\!\times\!\!\text{ 11=x}\left( \text{x+11} \right)$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+11x-1452=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+44x-33x-1452=0}$

$\Rightarrow \text{x}\left( \text{x+44} \right)\text{-33}\left( \text{x+44} \right)\text{=0}$

$\Rightarrow \left( \text{x+44} \right)\left( \text{x-33} \right)\text{=0}$

$\Rightarrow x\text{=-44 or 33}$

Since, speed cannot be negative.

Therefore, the speed of the passenger train will be $\text{33 km/h}$ and thus, the speed of the express train will be $\text{44 km/h}$ .

11. Sum of the areas of two squares is $\text{468 }{{\text{m}}^{\text{2}}}$. If the difference of their perimeters are $\text{24 m}$, find the sides of the two squares.

Ans: Let the sides of the two squares be $\text{x m}$ and $\text{y m}$.

Thus, their perimeters will be $\text{4x}$ and $\text{4y}$ and areas will be ${{\text{x}}^{2}}$ and ${{\text{y}}^{2}}$.

Hence, according to question –

$\text{4x-4y=24}$

$\Rightarrow \text{x-y=6}$

$\Rightarrow \text{x=y+6}$

And ${{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}}\text{=468}$

Substituting value of x-

${{\left( \text{y+6} \right)}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}}\text{=468}$

$\Rightarrow \text{36+}{{\text{y}}^{\text{2}}}\text{+12y+}{{\text{y}}^{\text{2}}}\text{=468}$

$\Rightarrow \text{2}{{\text{y}}^{\text{2}}}\text{+12y-432=0}$

$\Rightarrow {{\text{y}}^{\text{2}}}\text{+6y-216=0}$

$\Rightarrow {{\text{y}}^{\text{2}}}\text{+18y-12y-216=0}$

$\Rightarrow \text{y}\left( \text{y+18} \right)\text{-12}\left( \text{y+18} \right)\text{=0}$

$\Rightarrow \left( \text{y+18} \right)\left( \text{y-12} \right)\text{=0}$

$\Rightarrow \text{y=-18 or 12}$

Since, side cannot be negative.

Therefore, the sides of the square are $\text{12 m}$ and $\left( \text{12+6} \right)\text{m}$ i.e $\text{18 m}$.

Important Points from NCERT Class 10 Quadratic Equations

• A quadratic equation can be represented as:

ax2 + bx + c = 0

Where x is the variable of the equation and a, b and c are the real numbers. Also, a≠0.

• The nature of roots of a quadratic equation ax2 + bx + c = 0 can be find as:

• A real number α be root of quadratic equations ax2 + bx + c = 0 if and only if

aα2 + bα + c = 0.

Quadratic equations are very important in real-life situations. Learn all the concepts deeply and understand each topic conceptually. And, now let us solve questions related to quadratic equations.

Maths Class 10 Quadratic Equations Mind Map

Relation Between the Zeroes of a Quadratic Equation and the Coefficient of a Quadratic Equation

If α and β are zeroes of the quadratic equation $ax^2 + bx + c = 0$, where a, b, and c are real numbers and a ≠ 0, then

$\alpha + \beta = -\dfrac{b}{a}$

$\text{sum of zeros} = -\dfrac{\text{coefficient of x}}{\text{coefficient of }x^2}$

$\alpha \beta = \dfrac{c}{a}$

$\text{product of zeros} = -\dfrac{\text{constant term}}{\text{coefficient of }x^2}$

Methods of Solving a Quadratic Equation

The following are the methods that are used to solve quadratic equations:

(i) Factorization; (ii) Completing the Square; (iii) Quadratic Formula

Methods of Factorization

In this method, we find the roots of a quadratic equation $(ax^2 + bx + c = 0)$ by factorizing LHS into two linear factors and equating each factor to zero, e.g.,
$6x^2 - x - 2 = 0$
$\Rightarrow 6x^2 + 3x - 4x - 2 = 0$ …(i)
$\Rightarrow 3x (2x + 1) - 2(2x + 1) = 0$
$\Rightarrow (3x - 2) (2x + 1) = 0$
$\Rightarrow 3x - 2 = 0$ or $2x + 1 = 0$

Therefore $x = \dfrac{2}{3}$ or $x = -\dfrac{-1}{2}$

Method of Completing the Square

This is the method of converting the LHS of a quadratic equation that is not a perfect square into the sum or difference of a perfect square and a constant by adding and subtracting the terms.

Quadratic Formula

Consider a quadratic equation: ax2 + bx + c = 0.
If b2 – 4ac ≥ 0, then the roots of the above equation are given by:

$x = -\dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Overview of Deleted Syllabus for CBSE Class 10 Maths Chapter 4 Quadratic Equation

Class 10 Maths Chapter 4: Exercise Breakdown

Conclusion

NCERT Solutions for Class 10 Maths Chapter 4 - provides a comprehensive guide to understanding quadratic equations. This chapter is important for students because it teaches topics that are fundamental to advanced mathematics. The solutions describe how to solve quadratic equations, apply the quadratic formula, and investigate the nature of the roots using the discriminant. For effective exam preparation, focus on understanding formula derivation and applying the many types of problem-solving approaches described in this chapter. Last year, four to six questions from this area featured in the board exams, demonstrating its importance. These answers are precisely crafted to help students succeed by improving their problem-solving skills and conceptual understanding.

Other Study Materials of CBSE Class 10 Maths Quadratic Equation

Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.