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.

---

(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 \(\frac{2}{ \sqrt3}\) .

---

(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\(\frac{3-\sqrt 3}{2}\).