Find k for Equal Roots: Discriminant Trap | Class 10 Maths

Quick answer: A quadratic $ax^2 + bx + c = 0$ has two equal real roots when $D = b^2 - 4ac = 0$. If $k$ appears in the coefficient of $x^2$, reject any value of $k$ that makes that coefficient zero. For $(k - 12)x^2 + 2(k - 12)x + 2 = 0$, $D = 0$ gives $k = 12$ or $k = 14$, but $k = 12$ makes $a = 0$, so the answer is k = 14.

Discriminant and nature of roots

For the quadratic equation $ax^2 + bx + c = 0$ (with $a \ne 0$), the discriminant is:

$D = b^2 - 4ac$

So, for equal roots, set $b^2 - 4ac = 0$, that is, $b^2 = 4ac$.

The leading-coefficient trap

An equation $ax^2 + bx + c = 0$ is quadratic only if $a \ne 0$. If $a = 0$, the $x^2$ term disappears and the equation is no longer quadratic, so it cannot have "two equal roots".

Rule: After solving $D = 0$ for $k$, substitute each value back into the coefficient of $x^2$. Reject any value that makes it zero.

Problem 1: (k − 12)x² + 2(k − 12)x + 2 = 0

Find the value of $k$ for which this equation has two equal roots.

Step 1: Identify the coefficients.

$a = k - 12$, $b = 2(k - 12)$, $c = 2$

Step 2: Set the discriminant to zero.

$[2(k - 12)]^2 - 4(k - 12)(2) = 0$

$4(k - 12)^2 - 8(k - 12) = 0$

Step 3: Take out the common factor 4(k − 12).

Do not expand $(k - 12)^2$; factorising is quicker.

$4(k - 12)\,[(k - 12) - 2] = 0$

$4(k - 12)(k - 14) = 0$

So $k = 12$ or $k = 14$.

Step 4: Test each value in the coefficient of x².

• If $k = 12$: $a = 12 - 12 = 0$. The equation becomes $0 \cdot x^2 + 0 \cdot x + 2 = 0$, that is $2 = 0$, which is impossible. So reject k = 12.
• If $k = 14$: $a = 2 \ne 0$. The equation becomes $2x^2 + 4x + 2 = 0$, that is $x^2 + 2x + 1 = 0$, or $(x + 1)^2 = 0$. The roots are $-1$ and $-1$, which are equal. ✓

Answer: $k = 14$.

Problem 2: kx(x − 2) + 6 = 0 (NCERT)

Find the value of $k$ for which this equation has two equal roots.

Step 1: Write in standard form.

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

So $a = k$, $b = -2k$, $c = 6$.

Step 2: Set the discriminant to zero.

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

$4k^2 - 24k = 0$

$4k(k - 6) = 0$

So $k = 0$ or $k = 6$.

Step 3: Test each value.

• If $k = 0$: the equation becomes $6 = 0$, which is impossible. Reject k = 0.
• If $k = 6$: the equation is $6x^2 - 12x + 6 = 0$, that is $(x - 1)^2 = 0$, with equal roots $1$ and $1$. ✓

Answer: $k = 6$.

Summary table

Exam tips

• Always write the equation in standard form before reading off $a$, $b$ and $c$.
• Write the reason for rejecting a value in words, for example: "For the equation to be quadratic, the coefficient of $x^2$ cannot be zero, so $k = 12$ is rejected."
• Factorise instead of expanding when a bracket such as $(k - 12)$ is common to every term.
• Find the equal roots for your final $k$ as a quick check.

Common mistakes

• Giving both values ($k = 12$ or $k = 14$) as the answer without testing them.
• Dividing both sides by $(k - 12)$ without first stating $k \ne 12$. This can hide a solution or a rejected value.
• Forgetting to square the 2 in $b = 2(k - 12)$, which gives $2(k - 12)^2$ instead of $4(k - 12)^2$.
• Not expanding $kx(x - 2)$ before reading off the coefficients.

Frequently Asked Questions

What is the condition for equal roots of a quadratic equation?

The discriminant must be zero: b² − 4ac = 0. Both roots are then equal to −b/2a.

Why is k = 12 rejected in (k − 12)x² + 2(k − 12)x + 2 = 0?

Because k = 12 makes the coefficient of x² zero. The equation then reduces to 2 = 0, which is not a quadratic equation and has no roots at all.

What is the value of k if kx(x − 2) + 6 = 0 has equal roots?

k = 6. The discriminant condition gives k = 0 or k = 6, and k = 0 is rejected because it removes the x² term.

What happens when the discriminant is negative?

When b² − 4ac < 0, the quadratic equation has no real roots.

Can the coefficient of x² be zero in a quadratic equation?

No. If the coefficient of x² is zero, the equation becomes linear (or has no x term at all), so it is not a quadratic equation.