# Quadratic Equation

## 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.