Question

Find the nature of the roots of the following quadratic equations. If the real roots exist,
find them:
(ii) 3x2 -4 13x + 4 =0
(i) 2x2 -- 3x + 5 = 0
(iii) 2x2 - 6x + 3 = 0​

Answer 1

(i) Given, 2x2 – 3x + 5 = 0

Comparing the equation with ax2 + bx + c = 0, we get

a = 2, b = -3 and c = 5  

We know, Discriminant = b2 – 4ac

= ( – 3)2 – 4 (2) (5) = 9 – 40

= – 31

As you can see, b2 – 4ac < 0

Therefore, no real root is possible for the given equation, 2x2 – 3x + 5 = 0.

(ii) 3x2 – 4√3x + 4 = 0

Comparing the equation with ax2 + bx + c = 0, we get

a = 3, b = -4√3 and c = 4

We know, Discriminant = b2 – 4ac

= (-4√3)2 – 4(3)(4)

= 48 – 48 = 0

As b2 – 4ac = 0,  

Real roots exist for the given equation and they are equal to each other.

Hence the roots will be –b/2a and –b/2a.

–b/2a = -(-4√3)/2×3 = 4√3/6 = 2√3/3 = 2/√3

Therefore, the roots are 2/√3 and 2/√3.