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

For a quadratic equation \(ax^2 + bx + c = 0\), the discriminant is \(b^2 − 4ac\). The nature of the roots depends on the discriminant:

• If \(b^2 − 4ac>0\), there are two distinct real roots.
• If \(b^2 − 4ac = 0\), there are two equal real roots.
• If \(b^2 − 4ac<0\), there are no real roots.

(i) Consider the equation \(2x^2 −3x + 5 = 0\). Comparing with \(ax^2 + bx + c = 0\), we have a = 2, b = −3, c = 5. The discriminant is \(b^2 − 4ac = (− 3)^2 − 4(2)(5) = 9 – 40 = −31\). Since \(b^2 − 4ac<0\), this equation has no real roots.

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(ii) Consider the equation \(3x^2 -4\sqrt3 x +4 =0\). Comparing with \(ax^2 + bx + c = 0\), we have a = 3, b = \(4\sqrt3\), c = 4. The discriminant is \(b^2 -4ac = (-4\sqrt3)^2 -4(3)(4) = 48 − 48 = 0\). Since \(b^2 − 4ac = 0\), this equation has two equal real roots. The roots are given by \(-\frac{b}{2a}\). \(-\frac{b}{2a} = -\frac{(-4\sqrt3)}{ 2 \times 3 } = \frac{4\sqrt3}{6} = \frac{2\sqrt3 }{3} = \frac{2}{ \sqrt3}\). Therefore, the roots are \(\frac{2}{ \sqrt3}\) and \(\frac{2}{ \sqrt3}\).

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(iii) Consider the equation \(2x^2 − 6x + 3 = 0\). Comparing with \(ax^2 + bx + c = 0\), we have a = 2, b = −6, c = 3. The discriminant is \(b^2 − 4ac = (− 6)^2 − 4(2)(3) = 36 − 24 = 12\). Since \(b^2 − 4ac>0\), this equation has distinct real roots. The roots are given by \(x = -\frac{b±\sqrt{b^2 -4ac} }{ 2a}\). Substituting the values: \(x = -\frac{(-6) ± \sqrt{(-6)^2 - 4(2)(3)}}{ 2(2)} = \frac{6±\sqrt{12}}{4 } = \frac{6 ± 2\sqrt 3}{4} = \frac{3±\sqrt 3}{2}\). Therefore, the roots are \(\frac{3+\sqrt 3}{2}\) or \(\frac{3-\sqrt 3}{2}\).