Q2(ii):

Find the values of $k$ for each of the following quadratic equations, so that they have two equal roots. (ii) $kx (x – 2) + 6 = 0$

Given: A quadratic equation $kx(x - 2) + 6 = 0$, where $k \neq 0$.

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

Step 1: Simplifying the equation into standard form

The standard form of a quadratic equation is $ax^2 + bx + c = 0$. We expand the given equation:

$kx(x - 2) + 6 = 0$

$kx^2 - 2kx + 6 = 0$

[Distributive property of multiplication over subtraction]

Step 2: Identifying coefficients

Comparing $kx^2 - 2kx + 6 = 0$ with $ax^2 + bx + c = 0$, we identify:

$a = k$

$b = -2k$

$c = 6$

Step 3: Applying the condition for equal roots

For a quadratic equation to have two equal real roots, its discriminant ($D$) must be equal to zero.

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

Substituting the identified coefficients into the discriminant formula:

$(-2k)^2 - 4(k)(6) = 0$

Step 4: Solving for $k$

$4k^2 - 24k = 0$

[Simplifying the powers and products]

Factor out the common term $4k$:

$4k(k - 6) = 0$

[Using the zero-product property: if $p \cdot q = 0$, then $p=0$ or $q=0$]

Case 1: $4k = 0 \implies k = 0$

Case 2: $k - 6 = 0 \implies k = 6$

Step 5: Verification of validity

If $k = 0$, the equation becomes $0x^2 - 2(0)x + 6 = 0$, which simplifies to $6 = 0$. This is a contradiction and not a quadratic equation. Therefore, $k$ cannot be $0$.

If $k = 6$, the equation becomes $6x^2 - 12x + 6 = 0$. Dividing by 6, we get $x^2 - 2x + 1 = 0$, which is $(x - 1)^2 = 0$, yielding equal roots $x = 1, 1$.

Final Answer: The value of $k$ is 6.