Multiple choice

Find the value of $k$ for which the quadratic equation $\displaystyle \left ( k-2 \right )x^{2}+2\left ( 2k-3 \right )x+5k-6=0$ has equal roots

  1. 1

  2. 3

  3. A and B both

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For equal roots, the discriminant D = b^2 - 4ac = 0. Here a = k-2, b = 2(2k-3), c = 5k-6. D = 4(2k-3)^2 - 4(k-2)(5k-6) = 0. Simplifying: (4k^2 - 12k + 9) - (5k^2 - 16k + 12) = 0, which leads to -k^2 + 4k - 3 = 0, or k^2 - 4k + 3 = 0. Roots are k=1 and k=3.

AI explanation

For a quadratic equation to have equal roots, its discriminant must be zero, so we evaluate D = [2(2k - 3)]^2 - 4(k - 2)(5k - 6). Expanding the terms gives D = 4(4k^2 - 12k + 9) - 4(5k^2 - 16k + 12), which simplifies to 4(-k^2 + 4k - 3). Setting this equal to zero yields k^2 - 4k + 3 = 0, and factoring gives (k - 1)(k - 3) = 0. Therefore, the valid values are k = 1 and k = 3.

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