Roots of Quadratic Equations NCERT Exercise 4.3 part two Class Ten Mathematics

## Exercise 4.3 Part 2

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)