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) \(3x^2 - 4\sqrt{3}x + 4 = 0\)
(iii) \(2x^2 - 6x + 3 = 0\)
(i) Real roots do not exist.
(ii) Equal roots: \(\dfrac{2}{\sqrt{3}}, \dfrac{2}{\sqrt{3}}\)
(iii) Distinct roots: \(\dfrac{3 \pm \sqrt{3}}{2}\)
Key idea: For a quadratic equation \(ax^2 + bx + c = 0\), the discriminant is \(D = b^2 - 4ac\).
• If \(D > 0\): two distinct real roots.
• If \(D = 0\): two equal real roots.
• If \(D < 0\): no real roots (roots are non-real complex).
Step 1: Identify coefficients.
Here, \(a = 2,\ b = -3,\ c = 5\).
Step 2: Compute the discriminant.
\[D = b^2 - 4ac = (-3)^2 - 4(2)(5)\]
\[D = 9 - 40 = -31\]
Step 3: Interpret the discriminant.
Since \(D = -31 < 0\), the equation has no real roots. Its roots are non-real complex.
Conclusion for (i): Real roots do not exist.
Step 1: Identify coefficients.
Here, \(a = 3,\ b = -4\sqrt{3},\ c = 4\).
Step 2: Compute the discriminant.
\[D = b^2 - 4ac = (-4\sqrt{3})^2 - 4(3)(4)\]
First, \((-4\sqrt{3})^2 = 16 \cdot 3 = 48\).
So,
\[D = 48 - 4 \cdot 3 \cdot 4 = 48 - 48 = 0\]
Step 3: Interpret the discriminant.
Since \(D = 0\), the equation has two equal real roots.
Step 4: Find the equal root. For \(D = 0\), the root is
\[x = \dfrac{-b}{2a}\]
Substitute \(a = 3\) and \(b = -4\sqrt{3}\):
\[x = \dfrac{-(-4\sqrt{3})}{2 \cdot 3} = \dfrac{4\sqrt{3}}{6} = \dfrac{2}{\sqrt{3}}\]
So both roots are the same.
Conclusion for (ii): Equal real roots \(x = \dfrac{2}{\sqrt{3}}, \dfrac{2}{\sqrt{3}}\).
Step 1: Identify coefficients.
Here, \(a = 2,\ b = -6,\ c = 3\).
Step 2: Compute the discriminant.
\[D = b^2 - 4ac = (-6)^2 - 4(2)(3)\]
\[D = 36 - 24 = 12\]
Step 3: Interpret the discriminant.
Since \(D = 12 > 0\), the equation has two distinct real roots.
Step 4: Use the quadratic formula.
\[x = \dfrac{-b \pm \sqrt{D}}{2a} = \dfrac{-(-6) \pm \sqrt{12}}{2 \cdot 2}\]
\[x = \dfrac{6 \pm \sqrt{12}}{4}\]
Simplify \(\sqrt{12}\):
\[\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}\]
So,
\[x = \dfrac{6 \pm 2\sqrt{3}}{4} = \dfrac{6}{4} \pm \dfrac{2\sqrt{3}}{4} = \dfrac{3}{2} \pm \dfrac{\sqrt{3}}{2}\]
Combine over a common denominator:
\[x = \dfrac{3 \pm \sqrt{3}}{2}\]
Conclusion for (iii): Two distinct real roots: \(x = \dfrac{3 + \sqrt{3}}{2}\) and \(x = \dfrac{3 - \sqrt{3}}{2}\).