Ex 4.4 Q1. CBSE 10th Maths NCERT exercise solution

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

i. 2x2 – 3x + 5 = 0
ii. 3x2 - 4$$sqrt3$x + 4 = 0 iiii. 2x2 – 6x + 3 = 0 #### Explanatory Answer ##### i. 2x2 – 3x + 5 = 0 In this equation, a = 2, b = -3 and c = 5 ∴ Discriminant b2 – 4ac =$-3)2 – 4 × 2 × 5
= 9 – 40 = -36 < 0
The discriminant is negative.
So, the equation has NO real roots.

##### ii. 3x2 - 4 √3 x + 4 = 0

In this equation, a = 3, b = -4 √3 and c = 4
∴ discriminant b2 – 4ac = (-4 √3)2 – 4 × 3 × 4 = 48 – 48 = 0
Because the discriminant is zero, the equation has two equal real roots.
x = $$frac{-$-4$sqrt3$}{2 × 3}) = $\frac{4\sqrt3}{6}$ = $\frac{2\sqrt3}{3}$ = $\frac{2}{\sqrt{3}}$ The roots are $\frac{2}{\sqrt{3}}$ and $\frac{2}{\sqrt{3}}$ ##### iii. 2x2 – 6x + 3 = 0 In this equation a = 2, b = -6, c = 3 ∴ discriminant b2 - 4ac =$-6)2 – 4 × 2 × 3
= 36 – 24 = 12 > 0
The discriminant is positive.
So, the equation has two distinct real roots.
The roots are $$frac {-b\pm \sqrt {{b}^{2}-4ac}} {2a}$ x = $\frac {-$-6$$pm \sqrt {12}} {2 \times 2}$ = $\frac {6\pm \sqrt {12}} {4}$ = $\frac {6\pm 2\sqrt {3}} {4}$
x = $\frac{3 + \sqrt{3}}{2}$ and x = $\frac{3 - \sqrt{3}}{2}$