How to Find k When a Quadratic Equation Has Equal Roots

What condition gives equal roots?

A real quadratic equation ax² + bx + c = 0 has equal real roots when b² − 4ac = 0 and a ≠ 0. First identify the coefficients with their signs. Solve the resulting parameter equation, then discard values that make a zero.

Example: x² − 6x + k = 0

Here a = 1, b = −6 and c = k. The discriminant is (−6)² − 4 × 1 × k = 36 − 4k. Equal roots require 36 − 4k = 0, giving k = 9.

Substituting gives x² − 6x + 9 = (x − 3)². The repeated root is x = 3. A direct check produces 9 − 18 + 9 = 0. Finding k and finding the root are different steps; include whichever the question asks for.

Example with k in the leading coefficient

For kx² + 4x + 1 = 0, the discriminant is 16 − 4k. Setting it to zero gives k = 4, which preserves a nonzero leading coefficient. The equation becomes 4x² + 4x + 1 = (2x + 1)², so the repeated root is −1/2.

The nonzero-leading-coefficient check matters. In kx² + kx = 0, the discriminant is k². Setting it to zero gives k = 0, but then the equation is 0 = 0, not a quadratic with a repeated root. There is no valid k for equal quadratic roots in that example.

Common mistakes

Square the whole negative coefficient: (−6)² is 36. Do not set the discriminant greater than zero; that gives two distinct real roots. Also, equal roots need not be zero or positive.

Practice

1. x² + 8x + k = 0 has equal roots. k = 16, repeated root −4.
2. 9x² + kx + 1 = 0 has equal roots. k = 6 or −6; the repeated roots are −1/3 and 1/3 respectively.

Substitute each parameter back into the original equation before accepting it.

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