Q2(i):

Find the values of $k$ for each of the following quadratic equations, so that they have two equal roots. (i) $2x^2 + kx + 3 = 0$

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

To find: The value(s) of $k$ such that the given quadratic equation has two equal real roots.

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 $2x^2 + kx + 3 = 0$ with the standard form, we identify:

$a = 2$

$b = k$

$c = 3$

Step 2: State the condition for equal roots.

For a quadratic equation $ax^2 + bx + c = 0$, the nature of the roots is determined by the discriminant, denoted by $D$, where $D = b^2 - 4ac$.

The condition for a quadratic equation to have two equal real roots is that the discriminant must be equal to zero:

$D = b^2 - 4ac = 0$

Step 3: Substitute the coefficients into the discriminant formula.

Substituting $a = 2$, $b = k$, and $c = 3$ into the equation $b^2 - 4ac = 0$:

$(k)^2 - 4(2)(3) = 0$

Step 4: Solve the resulting equation for $k$.

Perform the multiplication inside the expression:

$k^2 - 24 = 0$

Isolate $k^2$ by adding 24 to both sides of the equation:

$k^2 = 24$

Take the square root of both sides to solve for $k$:

$k = \pm\sqrt{24}$

Simplify the radical expression $\sqrt{24}$:

$\sqrt{24} = \sqrt{4 \times 6} = \sqrt{4} \times \sqrt{6} = 2\sqrt{6}$

Therefore:

$k = \pm 2\sqrt{6}$

Step 5: Conclusion.

The values of $k$ for which the quadratic equation $2x^2 + kx + 3 = 0$ has two equal roots are $2\sqrt{6}$ and $-2\sqrt{6}$.

Final Answer: k = 2\sqrt{6}, -2\sqrt{6}