Quadratic Equation

NCERT Exercise 4.3

Question 2: Find the roots of the quadratic equations given in Q 1 above by applying the quadratic formula.

(i) `2x^2 – 7x + 3`

Solution: We have; a = 2, b = - 7 and c = 3

D can be calculated as follows:

`D=b^2-4ac`

`=(-7)^2-4xx2xx3`

`=49-25=25`

Now roots can be calculated as follows:

`α=(-b+sqrtD)/(2a)`

`=(7+sqrt(25))/(2xx2)=(7+5)/(4)`

`=(12)/(4)=3`

`β=(-b-sqrtD)/(2a)`

`=(7-5)/(4)=1/2`

(ii) `2x^2 + x – 4 = 0`

Solution: We have; a = 2, b = 1 and c = - 1

D can be calculated as follows:

`D=b^2-4ac`

`=1^2-4xx2xx(-4)`

`=1+32=33`

Now roots can be calculated as follows:

`α=(-b+sqrtD)/(2a)`

`=(-1+sqrt(33))/(4)`

`β=(-b-sqrtD)/(2a)`

`=(-1-sqrt(33))/(4)`

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

Solution: We have; `a = 4`, `b = 4sqrt3` and `c = 3`

D can be calculated as follows:

`D=b^2-4ac`

`=(4sqrt3)^2-4xx4xx3`

`=48-48=0`

Now; root can be calculated as follows:

`text(Root)=(-b)/(2a)`

`=(-4sqrt3)/(2xx4)=(-sqrt3)/(2)`