NCERT Solutions for Class 10 Maths Exercise 4.3 Chapter 4 Quadratic Equations – FREE PDF Download

NCERT Class 10 Maths Ch 4 is one of the most important ones in the NCERT syllabus. Duly following NCERT Solutions for Class 10 Maths Chapter
4 ensures you that there will be no hindrance when you opt for more advanced branches of Maths. This is where CoolGyan comes in. Our free Class 10 Quadratic Equations solutions will help you understand the chapter thoroughly.

NCERT Solutions for Class 10 Maths Chapter 4- Quadratic Equations

1. Find the roots of the following quadratic equations if they exist by the method of completing square.

(i)

(ii)

(iii)

(iv)

Ans. (i)

First we divide equation by 2 to make coefficient of x² equal to 1,

We divide middle term of the equation by 2x, we get

We add and subtract square of from the equation ,

⇒ $x^{2} + {(\frac{7}{4})}^{2} - \frac{7}{2} x + \frac{3}{2} - {(\frac{7}{4})}^{2} = 0$

⇒

=> ${(x - \frac{7}{4})}^{2} = \frac{25}{16}$

Taking Square root on both sides,

⇒

Therefore,

(ii)

Dividing equation by 2,

Following procedure of completing square,

⇒

Taking square root on both sides,

⇒

Therefore,

(iii)

Dividing equation by 4,

Following the procedure of completing square,

⇒

(iv)

Dividing equation by 2,

Following the procedure of completing square,

⇒

⇒ = $\frac{- 31}{16}$

Right hand side does not exist because square root of negative number does not exist.

Therefore, there is no solution for quadratic equation

2. Find the roots of the following Quadratic Equations by applying quadratic formula.

(i)

(ii)

(iii)

(iv)

Ans. (i)

Comparing quadratic equation with general form , we get a = 2, b = -7 and c = 3

Putting these values in quadratic formula

⇒

⇒ x = 3, ½

(ii)

Comparing quadratic equation with the general form , we get a = 2, b = 1 and c = −4

Putting these values in quadratic formula

⇒

(iii)

Comparing quadratic equation with the general form , we get a = 4, b = and c = 3

Putting these values in quadratic formula

⇒

A quadratic equation has two roots. Here, both the roots are equal.

Therefore,

(iv)

Comparing quadratic equation with the general form , we get a = 2, b = 1 and c = 4

Putting these values in quadratic formula

⇒

But, square root of negative number is not defined.

Therefore, Quadratic Equation has no solution.

3. Find the roots of the following equations:

(i)

(ii)

Ans. (i)

⇒

Comparing equation with general form ,

We get a = 1, b = −3 and c = −1

Using quadratic formula to solve equation,

$x = \frac{3 \pm \sqrt{{(- 3)}^{2} - 4 \times 1 \times (- 1)}}{2 \times 1}$

⇒

(ii)

⇒ $\frac{(x - 7) - (x + 4)}{(x + 4) (x - 7)} = \frac{11}{30}$

⇒ $\frac{- 11}{(x + 4) (x - 7)} = \frac{11}{30}$

⇒

Comparing equation with general form ,

We get a = 1, b = −3 and c = 2

Using quadratic formula to solve equation,

$x = \frac{3 \pm \sqrt{{(- 3)}^{2} - 4 (1) (2)}}{2 \times 1}$

⇒

⇒ x = 2, 1

4. The sum of reciprocals of Rehman’s ages (in years) 3 years ago and 5 years from now is $\frac{1}{3}$ . Find his present age.

Ans. Let present age of Rehman = x years

Age of Rehman 3 years ago = (x − 3) years.

Age of Rehman after 5 years = (x + 5) years

According to the given condition:

⇒

⇒ 3 (2x + 2) = (x − 3) (x + 5)

⇒ 6x + 6 =

⇒

Comparing quadratic equation with general form ,

We get a = 1, b = −4 and c = −21

Using quadratic formula

$x = \frac{4 \pm \sqrt{{(- 4)}^{2} - 4 (1) (- 21)}}{2 \times 1}$

⇒

⇒ x = 7, −3

We discard x=−3. Since age cannot be in negative.

Therefore, present age of Rehman is 7 years.

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

Ans. Let Shefali’s marks in Mathematics = x

Let Shefali’s marks in English = 30 − x

If, she had got 2 marks more in Mathematics, her marks would be = x + 2

If, she had got 3 marks less in English, her marks in English would be = 30 – x − 3 = 27 − x

According to given condition:

=> (x + 2) (27 − x) = 210

⇒ = 210

⇒

Comparing quadratic equation with general form ,

We get a = 1, b = −25 and c = 156

Applying Quadratic Formula

$x = \frac{25 \pm \sqrt{{(- 25)}^{2} - 4 (1) (156)}}{2 \times 1}$

⇒

⇒ x = 13, 12

Therefore, Shefali’s marks in Mathematics = 13 or 12

Shefali’s marks in English = 30 – x = 30 – 13 = 17

Or Shefali’s marks in English = 30 – x = 30 – 12 = 18

Therefore, her marks in Mathematics and English are (13, 17) or (12, 18).

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

Ans. Let shorter side of rectangle = xmetres

Let diagonal of rectangle = (x + 60) metres

Let longer side of rectangle = (x + 30) metres

According to pythagoras theorem,

=> =

⇒ =

⇒

Comparing equation with standard form ,

We get a = 1, b = −60 and c = −2700

Applying quadratic formula

$x = \frac{60 \pm \sqrt{{(- 60)}^{2} - 4 (1) (- 2700)}}{2 \times 1}$

⇒

⇒ x = 90, –30

We ignore –30. Since length cannot be in negative.

Therefore, x = 90 which means length of shorter side = 90 metres

And length of longer side = x + 30 = 90 + 30 = 120 metres

Therefore, length of sides are 90 and 120 in metres.

7. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Ans. Let the larger number be x, then square of smaller number be 8x and square of larger number be x².

According to condition:

$x^{2} - 8 x = 180$

$\Rightarrow$ $x^{2} - 8 x - 180 = 0$

$\Rightarrow$ $x^{2} - 18 x + 10 x - 180 = 0$

$\Rightarrow$ $x (x - 18) + 10 (x - 18) = 0$

$\Rightarrow$ $(x - 18) (x + 10) = 0$

$\Rightarrow$ $x - 18 = 0$ or $x + 10 = 0$

$\Rightarrow$ $x = 18$ $x = - 10$

When $x = 18$ , then square of smaller number = 144

Then smaller number = $\pm 12$

Therefore, two numbers are (12, 18) or (−12, 18)

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

Ans. Let the speed of the train = x km/hr

If, speed had been 5 km/hr more, train would have taken 1 hour less.

So, according to this condition

⇒

Comparing equation with general equation ,

We get a = 1, b = 5 and c = −1800

Applying quadratic formula

⇒

⇒ x = 40, −45

Since speed of train cannot be in negative. Therefore, we discard x = −45

Therefore, speed of train = 40 km/hr

9. Two water taps together can fill a tank in hours. The tap of larger diameter takes 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 time taken by tap of smaller diameter to fill the tank = x hours

Let time taken by tap of larger diameter to fill the tank = (x – 10) hours

It means that tap of smaller diameter fills part of tank in 1 hour.… (1)

And, tap of larger diameter fills part of tank in 1 hour. … (2)

When two taps are used together, they fill tank in $\frac{75}{8}$ hours.

In 1 hour, they fill part of tank… (3)

From (1), (2) and (3),

⇒

⇒ 75 (2x − 10) =

⇒ 150x – 750 =

⇒

Comparing equation with general equation ,

We get a = 4, b = −115 and c = 375

Applying quadratic formula

⇒

⇒ x = 25, 3.75

Time taken by larger tap = x – 10 = 3.75 – 10 = −6.25 hours

Time cannot be in negative. Therefore, we ignore this value.

Time taken by larger tap = x – 10 = 25 – 10 = 15 hours

Therefore, time taken by larger tap is 15 hours and time taken by smaller tap is 25 hours.

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

Ans. Let average speed of passenger train = x km/h

Let average speed of express train = (x + 11) km/h

Time taken by passenger train to cover 132 km = hours