## NCERT Exercise 4.3

### Part 1

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

(i) 2x^2 – 7x + 3 = 0

Solution: 2x^2 – 7x + 3 = 0

Checking the existence of roots:

We know;

D=b^2-4ac
=(-7)^2-4xx2xx3
=49-24=25

Since D > 0; hence two different roots are possible for this equation.

Now; 2x^2 – 7x + 3 can be written as follows:

x^2-(7)/(2)x+3/2=0

Or, x^2-2(7/4)x+3/2=0

Or, x^2-2(7/4)x=-3/2

Or, x^2-2(7/4)x+(7/4)^2=(7/4)^2-3/2

Assuming x = a and 7/4 = b, the LHS of equation is made in the form of (a – b)2

Or, (x-7/4)^2=(49)/(16)-3/2

Or, (x-7/4)^2=(49-24)/(16)

Or, (x-7/4)^2=(25)/(16)

Case 1:

Or, x-7/4=5/4

Or, x=5/4+7/4=(12)/(4)=3

Case 2:

Or, x-7/4=-5/4

Or, x=-5/4+7/4

Or, x=(-5+7)/(4)=2/4=1/2

Hence; x = 3 and x =1/2

(ii) 2x^2 + x – 4 = 0

Solution: Checking the existence of roots:

We know;

D=b^2-4ac

=1^2-4xx2xx(-4)

=1+32=33

Since D > 0; hence roots are possible for this equation.

By dividing the equation by 2; we get following equation:

x^2+x/2-2=0

Or, x^2+2(1/4)x-2=0

Or, x^2+2(1/4)x=2

Or, x^2+2(1/4)x+(1/4)^2=2+(1/4)^2

Assuming x = a and 1/4 = b, the above equation can be written in the form of (a + b)2

Or, (x+1/4)^2=2+(1)/(16)

Or, (x+1/4)^2=(33)/(16)

Or, x+1/4=±(sqrt(33))/(16)

Case 1:

x=sqrt(33)/(16)-1/4

=(sqrt(33)-4)/(16)

Case 2:

x=-sqrt(33)/(16)-1/4

=(-sqrt(33)-4)/(16)

(iii) 4x^2 + 4sqrt3x + 3 = 0

Solution: Checking the existence of roots:

We know;

D=b^2-4ac

=(4sqrt3)^2-4xx4xx3

=48-48=0

Since D = 0; hence roots are possible for this equation.

After dividing by 4; the equation can be written as follows:

x^2+sqrt3x+3/4=0

Or, x^2+2(sqrt3/2)x=-3/2

Or, x^2+2(sqrt3/2)x+(sqrt3/2)^2=(sqrt3/2)^2-3/4

LHS of equation can be written in the form of (a + b)2:

Or, (x+sqrt3/2)^2=3/4-3/4=0

Or, x+sqrt3/2=0

Or, x=-sqrt3/2

(iv) 2x^2 + x + 4

Solution: Checking the existence of roots:

We know;

D=b^2-4ac
=1^2-4xx2xx4=-31

Since D < 0; hence roots are not possible for this equation.