Ncert Solutions For Class 10 Maths Ex 4.2

Ncert Solutions For Class 10 Maths Chapter 4 Ex 4.2

Q1. Find the roots of quadratic equation by factorization:

x2(3+2)x+6 = 0

 Sol.

x23x2x+6=0

 

x(x3)2(x3)=0

 

(x3)(x2)=0

 

Therefore x=2   or   x=3

 

Q2. Find the roots of quadratic equation:

10x273x+3=0

 Sol.

10x273x+3=0

 

x273x10+310=0

 

x2(32+35)x+310=0

 

x232x35x+310=0

 

x(x32)35(x32)=0

 

(x32)(x35)=0

 

Therefore   x=32   or   x=35

 

Q3. Find the roots of quadratic equation:

                   400x2 – 40x + 1 = 0

 Sol.

Since,   400x2 – 40x + 1 = 0

Therefore, x2110x+1400=0

 

x2110x+1400=0

 

x2(120+120)x+1400=0

 

x2120x120x+1400=0

 

x(x120)120(x120)

 

(x120)(x120)

 

Therefore          x=120   or   x=120

 

Q4. Find the roots of quadratic equation:

                      x2– 16x + 63 = 0

 Sol.

Since,  x2– 16x + 63 = 0

Therefore,   x2– (9+7)x + 63 = 0

x2– 9x-7x + 63 = 0

x(x –  9) – 7( x – 9 ) = 0

(x -9) ( x – 7 ) = 0

Therefore,  x = 9  or  x = 7

 

Q5. Find the roots of quadratic equation:

3x2+7x+23=0

 Sol.

3x2+7x+23=0

 

x2+73x+2=0

 

x2+63x+13x+2=0

 

x(x+63)+13(x+23)=0

 

(x+13)(x+23)=0

 

Therefore x=13   or   x=23

 

Q.6 Product of two numbers x and y is 144 and their sum is 25. Find x and y.

Sol.

According to given condition:

 x . y = 144 ……………………….. (1)

And,

 x + y = 25 ……………………………. (2)

Substituting equation (2) in (1) we get:

x(25 – x) = 144

(or)

x2 – 25x + 144= 0

x2 – (16 + 9)x + 144= 0

x2 – 16x – 9x + 144= 0

x(x – 16) – 9(x – 16) = 0

Therefore, x = 16 or   x = 9

So, if first number (x) is 16 then second number (y) will be (25 -16) = 9

(or)

If first number is(x) 9 then second number(y) will be (25-9) = 16.

 

Q7. Sum of square of two consecutive positive integers is 481, find the numbers.

Sol.

Let,  1st number be x

Therefore,     second number will be (x+1)

According to the given condition:

x2 + (x + 1)2 = 481

x2 + x2 + 2x + 1 = 481

2x2 + 2x – 480 = 0

x2 + x – 240 = 0

x2 + (16 – 15) x – 240 = 0

x2 + 16x – 15x – 240 = 0

x(x + 16) – 15(x + 16) = 0

(x – 15) (x + 16) = 0

Therefore,

x= 15   (or)   x=  -16

Since, the positive integers can’t be -16

Hence two consecutive positive integers are 15 and 16.

 

Q.8 The base of a right angled triangle is 3cm more than its altitude. If the hypotenuse is 15cm. Find the length of its base and altitude.

Sol.

In right angled triangle ABC:

Capture1

Let,        length of altitude = x cm

Therefore                   base = (x + 3) cm

Now, according to the given condition:

(AB)+ (AC)2 = (BC)2   (From Pythagoras theorem)

(x)+ (x+3)2 = (15)2

x+ x2 +6x + 9 = 225

2x2 + 6x – 216 = 0

x2 + 3x – 108 = 0

This is required quadratic equation, on further solving this equation:

x2 + 3x – 108 = 0

x2 + (12 – 9)x – 108 = 0

x2 + 12x – 9x – 108 = 0

x(x + 12) – 9(x + 12) = 0

(x + 12) (x – 9) = 0

Therefore, x = -12 (or) x = 9

Since length cannot be negative. Therefore, neglecting -12cm

Hence the length of altitude = 9 cm and length of the base = 12cm

 Finding roots by method of completing the square

Equations like x– 4x – 8 = 0, cannot be solved by method of factorization because it doesn’t factor.

These types of equations can be solved by method of completing the square.

i.e       x2 – 4x – 8 = 0   or  x2 – 4x = 8

Our main target is to convert x2  –  4x in form (x – a)2 which is equal to x2 – 2ax + a2.

So, we will try to break the middle term in such a way, that 4x can be written in form of 2ax.

i.e   4x= 2(2.x) thus, here a = 2.

Therefore, if we will add 22 on both LHS and RHS, we get the following equation:-

x2 – 2(2x) + 22 = 8 +22

(or),   (x – 2)2 = 12

(x – 2)2 = 12

(x – 2)2(12)2= 0

Since,           a2b2=(a+b)(ab)

Therefore,  (x212)  (x2+12)

x=2+12   or   x=212

This is how we find roots of a given equation by method of completing the square.