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

i) 2x^{2} - 3x + 5 = 0 ii) 3x^{2} - 4√3 x + 4 = 0

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

**Solution:**

The general form of a quadratic equation is ax^{2} + bx + c = 0

b^{2} - 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 b^{2} - 4ac > 0

(ii) Two equal real roots, if b^{2} - 4ac = 0

(iii) No real roots, if b^{2} - 4ac < 0

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

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

b^{2} - 4ac = (- 3)^{2} - 4 (2) (5)

= 9 - 40

= - 31

b^{2} - 4ac < 0

Hence the equation has no real roots.

(ii) 3x^{2} - 4√3 x + 4 = 0

a = 3, b = - 4√3, c = 4

b^{2} - 4ac = (- 4√3)^{2} - 4(3)(4)

= 16 × 3 - 4 × 4 × 3

= 48 - 48

= 0

b^{2} - 4ac = 0

Hence the equation has two equal real roots.

We know that, x = [- b ± √ (b^{2} - 4ac)] / 2a

x = - b/2a [Since, b^{2} - 4ac = 0]

x = -(- 4√3)/2(3)

= 2/√3

Roots are 2/√3, 2/√3

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

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

b^{2} - 4ac = (- 6)^{2} - 4(2)(3)

= 36 - 24

= 12

b^{2} - 4ac > 0

Hence the equation has two distinct real roots.

We know that, x = [- b ± √ (b^{2} - 4ac)] / 2a

x = [-(- 6) ± √12] / (2)2

= (6 ± 2√3) / 4

= (3 ± √3) / 2

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

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 4

**Video Solution:**

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

Class 10 Maths NCERT Solutions Chapter 4 Exercise 4.4 Question 1

**Summary:**

The nature of the roots of the following quadratic equations are such that i) 2x^{2} - 3x + 5 = 0 has no real roots, ii) 3x^{2} - 4√3x + 4 = 0 has two equal rooots 2/√3, 2/√3 and iii) 2x^{2} - 6x + 3 = 0 has two distinct real roots (3 + √3)/2, (3 - √3)/2 respectively.

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