NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equations

 NCERT Solutions For Class 10 Maths Chapter 4 Quadratic Equation

Exercise 4.1

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

(i) The area of a rectangular plot is `528 m^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.

Solution:

Let the beadth be `x`.

Then, the length is `2x + 1`

Area of the rectangular plot is `528 m^2`

Area of the rectangular plot = `l × b = 528`

`⇒ x (2x + 1) = 528`

`⇒ 2x^2 + x = 528`

`⇒ 2x^2 + x -528 = 0`

Required representation of the problem mathematically. 

(ii) The product of two consecutive positive integers is 306. We need to find the integers.

Solution:

Let the one positive integer be `x`.

Then, another positive integer is `= x+1`.

The product of two consecutive positive integers is 306.

`x(x+1) = 306`

`⇒ x^2 + x = 306`

`⇒ x^2 + x - 306 = 0`

Mathematical representation of the problem. 

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

Solution: 

Let the Rohan's age be `x` years.

Then, the Rohan's mother age `= x+ 26`.

3 years from now.

Rohan's age `= x+ 3`

Rohan's Mother age `= x + 26 + 3 = x + 29`

The product of their ages (in years) 3 years from now will be 360.

`(x+ 3) (x + 29) = 360`

`⇒ x^2 + 29x + 3x + 87 = 306`

`⇒   x^2 + 29x + 3x + 87 = 306`

`⇒ x^2 + 32x - 273`

Mathematical representation of the problem. 

  

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

Solution: 

Let the speed of the train be `x` km/h.

Distance traveled `= 480 km`.

Time taken  = `480/x`

If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance.

Speed `= x -8`

time taken = `480/x + 3`

`(x - 8)(48/x + 3) = 480`

`⇒ 480 + 3x -3840/x  - 24 = 480`

Work in progress 

Exercise 4.2

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

(i)  `x^2 – 3x – 10 = 0`

`⇒ x^2 -5x + 2x -10 = 0`

`⇒ x(x - 5) + 2 (x- 5) = 0`

`⇒ (x - 5) = 0` or `(x + 2) = 0`

 `x = 5` or `x = -2`

(ii) `2x^2 + x – 6 = 0`

`⇒ 2x^2 +4x -3x -6 = 0`

`⇒ 2x(x + 2)  -3(x+ 2)= 0`

`⇒ (x + 2) = 0` or `2x-3 = 0`

`x = -2` or `x = 3/2`

(iii) `sqrt2 x^2 + 7 x + 5 sqrt2 = 0`

`⇒ sqrt2 x^2 + 5x + 2x + 5 sqrt2 = 0`

`⇒ x(sqrt2 x + 5)  + sqrt2(sqrt2 x + 5) = 0` 

`⇒ sqrt2 x + 5`  = 0 or `(x + sqrt2) = 0`

`⇒ x = -5/sqrt2` or `x = -sqrt2`

(iv)  `2x^2 – x +1/8 = 0`

`⇒ 1/8 (16x^2 - 8x + 1) = 0`

`⇒ 16x^2 - 4x - 4x+ 1 = 0`

`⇒ 4x(4x -1)  - 1(4x -1) = 0`

`⇒ (4x -1)^2 = 0`

`x = 1/4` or `x = 1/4`

(v) `100x^2 – 20x + 1 = 0`

 `⇒ 100x^2 – 10x - 10x + 1 = 0`

`⇒ 10x(10x -1)  -1(10x - 1) = 0`

`⇒ (10x -1)^2 = 0`

 `x = 1/10` or `x = 1/10`

Work in progress 

Exercise 4.4

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

(i) `2x^2 – 3x + 5 = 0`

Solution:

`a = 2, b = -3, c = 5`

`D = b^2 - 4ac`

`= (-3)^2 - 4 × 2 × 5`

`= 9 - 40`

`= -31` 

`D ＜ -31`

So, no real root is exist. 

(ii) `3x^2 – 4 sqrt3 x + 4 = 0`

 `a= 3, b = -4sqrt3, c= 4`

 `D =  b^2 - 4ac`

`= (-4sqrt3)^2 - 4× 3 × 4`

`= 48 - 48 = 0`

`D = 0`

So, real roots exist, and they are equal.

The roots are `frac{-b}{2a}, frac{-b}{2a}`, i.e, `-frac{(-4sqrt3)}{6},-frac{(-4sqrt3)}{6}` 

i.e, `(2sqrt3)/3, (2sqrt3)/3`

(iii) `2x^2 – 6x + 3 = 0`

`a = 2 , b = -6, c = 3`

`D =  b^2 - 4ac`

`=  (-6)^2 -4 × 2 × 3`

`= 36 - 24`

`D = 12`

So, real roots exist and they are distinct.

 The roots are

` frac{-b ± sqrtD}{2a}`

`= -frac{(-6) ± sqrt12 }{2× 2 }`

`= frac{6 ± 2sqrt3 }{4}`

`= frac{3 ± sqrt3}{2}`

2. Find the values of k for each of the following quadratic equations, so that they have two equal roots.

(i) `2x^2 + kx + 3 = 0`

`a = 2, b = k, c = 3`

They have equal roots

So, `D = 0`

`b^2 -4ac = 0`

` ⇒ k^2 - 4×2×3 = 0`

`⇒ k^2 - 24 = 0`

`⇒ k =  sqrt24`

`⇒ k = ±2sqrt6`

(ii) `kx (x – 2) + 6 = 0`

`⇒ kx^2 - 2k + 6 = 0`

`a = k, b = -2k, c = 6`

They have equal roots

So, `D = 0`

`b^2 -4ac = 0`

` ⇒ (-2k)^2 - 4× k× 6 = 0`

` ⇒ 4k^2 - 24k = 0`

` ⇒ 4k(k- 6) = 0`

` ⇒k = 6`

3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is `800 m^2?` If so, find its length and breadth.

Solution:

Let the breadth be `x`.

Length is `2x`

area `= l×b = 800`

`2x^2 = 800`

`⇒ x^2 = 400`

`⇒ x = sqrt400`

`⇒ x = 20`

breadth `= 20`m

length `= 40`m

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

Solution:

Let the age of one friend be `x`, 

then the age of the other friend is `20 -x`

Four years ago 

`x -4  &  16 -x`

The product of their ages is 48

`(x-4)(16-x) = 48`

`⇒ 16x - x^2 - 64 + 4x = 48`

`⇒ - x^2 + 20x -112 = 0`

`a = -1, b = 20, c = -112`

`D = b^2 - 4ac`

= 400 - 4× (-1)×(-112)

`= 400 - 448`

`D= -48`

`D < 0`

So, no real root is exist.