Question 1

We will use the discriminant to determine the nature of the roots.

Step 1 — Analyze Equation (i)

Let's look at the first equation. The equation is $2x^2 - 3x + 5 = 0$. We compare it to $ax^2 + bx + c = 0$. Here, a = 2, b = -3, and c = 5. Now, we calculate the discriminant D.

$D = b^2 - 4ac$

$D = (-3)^2 - 4 \times 2 \times 5$

$D = 9 - 40$

$\boxed{D = -31}$

Since D < 0, the roots are not real. There are no real roots for this equation.

Step 2 — Analyze Equation (ii)

Let's look at the second equation. The equation is $3x^2 - 4\sqrt{3} x + 4 = 0$. We compare it to $ax^2 + bx + c = 0$. Here, a = 3, b = -4√3, and c = 4. Now, we calculate the discriminant D.

$D = b^2 - 4ac$

$D = (-4\sqrt{3})^2 - 4 \times 3 \times 4$

$D = (16 \times 3) - 48$

$D = 48 - 48$

$\boxed{D = 0}$

Since D = 0, the roots are real and equal. We use the quadratic formula to find them.

$x = \frac{-b \pm \sqrt{D}}{2a}$

$x = \frac{-(-4\sqrt{3}) \pm \sqrt{0}}{2 \times 3}$

$x = \frac{4\sqrt{3}}{6}$

$x = \frac{2\sqrt{3}}{3}$

$\boxed{\text{Roots are } \frac{2\sqrt{3}}{3}, \frac{2\sqrt{3}}{3}}$

Step 3 — Analyze Equation (iii)

Let's look at the third equation. The equation is $2x^2 - 6x + 3 = 0$. We compare it to $ax^2 + bx + c = 0$. Here, a = 2, b = -6, and c = 3. Now, we calculate the discriminant D.

$D = b^2 - 4ac$

$D = (-6)^2 - 4 \times 2 \times 3$

$D = 36 - 24$

$\boxed{D = 12}$

Since D > 0, the roots are real and distinct. We use the quadratic formula to find them.

$x = \frac{-b \pm \sqrt{D}}{2a}$

$x = \frac{-(-6) \pm \sqrt{12}}{2 \times 2}$

$x = \frac{6 \pm 2\sqrt{3}}{4}$

$x = \frac{2(3 \pm \sqrt{3})}{4}$

$x = \frac{3 \pm \sqrt{3}}{2}$

$\boxed{\text{Roots are } \frac{3 + \sqrt{3}}{2}, \frac{3 - \sqrt{3}}{2}}$

Answer

(i) The roots are not real. (ii) The roots are real and equal: $\frac{2\sqrt{3}}{3}, \frac{2\sqrt{3}}{3}$. (iii) The roots are real and distinct: $\frac{3 + \sqrt{3}}{2}, \frac{3 - \sqrt{3}}{2}$.