Question

Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

(i) \( 2 x^{2}-3 x+5=0 \)

(ii) \( 3 x^{2}-4 \sqrt{3} x+4=0 \)

(iii) \( 2 x^{2}-6 x+3=0 \)

Answer 1

The general form of a quadratic equation is ax² + bx + c = 0

b² - 4ac is called the discriminant of the quadratic equation and we can decide whether the real roots exist or not based on the value of the discriminant.

We know that,

(i) Two distinct real roots, if b² - 4ac > 0

(ii) Two equal real roots, if b² - 4ac = 0

(iii) No real roots, if b² - 4ac < 0

(i) 2x² - 3x + 5 = 0

a = 2 , b = -3, c = 5

b² - 4ac = (- 3)² - 4 (2) (5)

= 9 - 40

= - 31

b² - 4ac < 0

Hence the equation has no real roots.

(ii) 3x² - 4√3 x + 4 = 0

a = 3, b = - 4√3, c = 4

b² - 4ac = (- 4√3)² - 4(3)(4)

= 16 × 3 - 4 × 4 × 3

= 48 - 48

= 0

b² - 4ac = 0

Hence the equation has two equal real roots.

We know that, \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

x = - b/2a [Since, b² - 4ac = 0]

x = -(- 4√3)/2(3)

= 2/√3

Roots are 2/√3, 2/√3

(iii) 2x² - 6x + 3 = 0

a = 2, b = - 6, c = 3

b² - 4ac = (- 6)² - 4(2)(3)

= 36 - 24

= 12

b² - 4ac > 0

Hence the equation has two distinct real roots.

We know that, \(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

x = [-(- 6) ± √12] / (2)2

= (6 ± 2√3) / 4

= (3 ± √3) / 2

Roots are x = (3 + √3) / 2 and x = (3 - √3) / 2