Question

Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:
(i) $2x^2 – 3x + 5 = 0$ (ii) $3x^2 – 4\sqrt3 x + 4 = 0$ (iii) $2x^2 – 6x + 3 = 0$

Answer 1

We know that for a quadratic equation $ax^2 + bx + c = 0$, discriminant is $b^2 − 4ac$.

• If $b^2 − 4ac$ > 0 → two distinct real roots
• If $b^2 − 4ac$ = 0 → two equal real roots
• If $b^2 − 4ac$ < 0 → no real roots

(i) $2x^2 −3x + 5 = 0$
Comparing this equation with $ax^2 + bx + c = 0$, we obtain
a = 2, b = −3, c = 5

Discriminant = $b^2 − 4ac$ =$(− 3)2 − 4 (2) (5)$= $9 – 40$ = $−31$
As $b^2 − 4ac < 0$, Therefore, no real root is possible for the given equation.

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**(ii) **$3x^2 -4\sqrt3 x +4 =0$
Comparing this equation with $ax^2 + bx + c = 0$, we obtain
a=$3$ , b= $4\sqrt3$, c=$4$

Discriminant =$b^2 -4ac$ = $(-4\sqrt3)^2 -4(3)(4)$ = $48 − 48$ = 0
As $b^2 − 4ac$ = 0, Therefore, real roots exist for the given equation and they are equal to each other.

And the roots will be $−\frac{𝑏}{2𝑎}$ and $−\frac{𝑏}{2𝑎}$ .
$−\frac{𝑏}{2𝑎}$= $-\frac{(-4\sqrt3)}{ 2 \times 3 }$=$\frac{ 4\sqrt3}{6}$ = $\frac{2\sqrt3 }{3}$ = $\frac{2}{ \sqrt3}$

Therefore, the roots are $\frac{2}{ \sqrt3}$ and $\frac{2}{ \sqrt3}$ .

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(iii) $2x^2 − 6x + 3 = 0$
Comparing this equation with $ax^2 + bx + c = 0$, we obtain
a = 2, b = −6, c = 3

Discriminant = $b^2 − 4ac$= $(− 6)2 − 4 (2) (3)$= $36 − 24$ = $12$
As b^2 − 4ac$$> 0, Therefore, distinct real roots exist for this equation as follows.

x= $-\frac{b±\sqrt{b^2 -4ac} }{ 2a}$
= $-\frac{(-6) ± \sqrt{(-6)^2 - 4(2)(3)}}{ 2(2)}$
= $\frac{6±\sqrt{12}}{4 }$= $\frac{6 ± 2\sqrt 3}4$
= $\frac{3±\sqrt 3}{2}$

Therefore, the roots are $\frac{3+\sqrt 3}{2}$ or$\frac{3-\sqrt 3}{2}$.