Q1(iii):

Find the nature of the roots of the following quadratic equations. If the real roots exist, find them: (iii) $2x^2 – 6x + 3 = 0$

Given: A quadratic equation $2x^2 - 6x + 3 = 0$.

To Find: The nature of the roots of the given quadratic equation and, if real roots exist, find their values.

Step 1: Identify the coefficients of the quadratic equation.

A standard quadratic equation is represented as $ax^2 + bx + c = 0$. Comparing the given equation $2x^2 - 6x + 3 = 0$ with the standard form:

• $a = 2$
• $b = -6$
• $c = 3$

Step 2: Determine the nature of the roots using the Discriminant ($D$).

The discriminant of a quadratic equation is given by the formula $D = b^2 - 4ac$.

Substituting the values of $a$, $b$, and $c$:

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

$D = 36 - 24$

$D = 12$

[Since $D > 0$, the quadratic equation has two distinct real roots.]

Step 3: Apply the Quadratic Formula to find the roots.

The quadratic formula is given by $x = \frac{-b \pm \sqrt{D}}{2a}$.

Substituting the known values into the formula:

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

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

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

Step 4: Simplify the expression.

Factor out the common term $2$ from the numerator:

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

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

Therefore, the two roots are:

$x_1 = \frac{3 + \sqrt{3}}{2}$

$x_2 = \frac{3 - \sqrt{3}}{2}$

Final Answer: The roots are real and distinct. The roots of the equation are $\frac{3 + \sqrt{3}}{2}$ and $\frac{3 - \sqrt{3}}{2}$.