Find the nature of the roots of the following . If the real roots exist, find them: (i) 2x2 – 3x + 5 = 0 (ii) 3x2 – 4 3 x + 4 = 0 (iii) 2x2 – 6x + 3 = 0
Step-by-Step Solution
Key Concept: For a quadratic equation $ax^2+bx+c=0$, the discriminant $D=b^2-4ac$ determines the nature of its roots: <br>• $D>0$ – two distinct real roots, <br>• $D=0$ – equal real roots, <br>• $D<0$ – no real roots (complex conjugate roots). Real roots, when they exist, are given by $x=\frac{-b\pm\sqrt{D}}{2a}$.
1. Equation (i): $2x^2-3x+5=0$
• Identify $a=2$, $b=-3$, $c=5$.
• Compute discriminant: $$D = b^2-4ac = (-3)^2-4\times2\times5 = 9-40 = -31.$$
• Since $D<0$, the equation has no real roots; the roots are complex conjugates.
2. Equation (ii): $3x^2-4\sqrt{3}\,x+4=0$
• Identify $a=3$, $b=-4\sqrt{3}$, $c=4$.
• Compute discriminant: $$D = b^2-4ac = (-4\sqrt{3})^2-4\times3\times4 = 16\times3-48 = 48-48 = 0.$$
• Since $D=0$, the equation has equal real roots.
• Root: $$x = \frac{-b}{2a}=\frac{4\sqrt{3}}{2\times3}=\frac{2\sqrt{3}}{3}.$$
• Hence the double root is $x=\frac{2\sqrt{3}}{3}$.
3. Equation (iii): $2x^2-6x+3=0$
• Identify $a=2$, $b=-6$, $c=3$.
• Compute discriminant: $$D = b^2-4ac = (-6)^2-4\times2\times3 = 36-24 = 12.$$
• Since $D>0$, the equation has two distinct real roots.
• Roots: $$x = \frac{-b\pm\sqrt{D}}{2a}=\frac{6\pm\sqrt{12}}{4}=\frac{6\pm2\sqrt{3}}{4}=\frac{3\pm\sqrt{3}}{2}.$$
• Therefore the roots are $x_1=\frac{3+\sqrt{3}}{2}$ and $x_2=\frac{3-\sqrt{3}}{2}$.
Correct Answer: (i) No real roots (complex conjugate).\n(ii) Equal real roots: $x=\frac{2\sqrt{3}}{3}$ (double root).\n(iii) Two distinct real roots: $x=\frac{3+\sqrt{3}}{2}$ and $x=\frac{3-\sqrt{3}}{2}$.