Question

Please can anyone help me to solve this problem:'(​

Answer 1

Question

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

a) 2x² - 3x + 5 = 0

b) 3x² - 4√3x + 4 = 0

c) 2x² - 6x + 3 = 0

Solution

The nature of the roots (real distinct, real equal or imaginary) is obtained by finding the discriminant (b² - 4ac) of the equation. If :-

(b² - 4ac) > 0, then real distinct roots

(b² - 4ac) = 0, then real equal roots

(b² - 4ac) < 0, then imaginary roots.

For the first equation

2x² - 3x + 5 = 0,

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

So, b² - 4ac = (-3)² - 4(2)(5)

→ 9 - 40

= -31

Now, -31 < 0

→ b² - 4ac < 0

Hence, the nature of the roots are imaginary.

$\rule{200}1$

b) 3x² - 4√3x + 4 = 0

b = -4√3

a = 3

c = 4

b² - 4ac = (-4√3)² - 4(3)(4)

→ b² - 4ac = 48 - 48

→ b² - 4ac = 0

Hence, the roots are real and equal. Now, we will have to find the roots also

Factorising it, we get

3x² - 2√3x - 2√3x + 4 = 0

→ √3x(√3x - 2) - 2(√3x - 2) = 0

→ (√3x - 2)(√3x - 2) = 0

→ √3x - 2 = 0

→ x = 2/(√3)

→ x = 2√3/3

Root of the equation is 2√3/3

$\rule{200}1$

c) 2x² - 6x + 3 = 0

b = - 6

a = 2

c = 3

b² - 4ac = (-6)² - 4(2)(3)

→ b² - 4ac = 36 - 24

→ b² - 4ac = 12

Again, the roots of this equation are real and distinct

So to find the roots let's apply quadratic formula

$\sf x = \frac{-b \pm\sqrt{b^2 - 4ac}}{2a}$

$\sf x = \frac{-b\pm \sqrt{12}}{2a}$

since, b² - 4ac = 12

$\sf x = \frac{-(-6) + \sqrt{12}}{2\times 2} \ or, x = \frac{-(-6)-\sqrt{12}}{2\times 2}\\ \\ \implies x = \frac{6 + 2\sqrt{3}}{4} \ or, x =\frac{6 - 2\sqrt{3}}{4}\\ \\ \implies x = \frac{3 + \sqrt{3}}{2} \ or, x = \frac{3 - \sqrt{3}}{2}$

Hence the roots of the equation are (3 + √3)/2 or (3 - √3)/2