Multiple choice

The value of k for which the roots are real and equal of the following equation $k^2x^2$ - 2(2k - 1) x + 4 = 0 is k = $\dfrac{1}{4}$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For real and equal roots, the discriminant D = b^2 - 4ac = 0. Here, a = k^2, b = -2(2k-1), c = 4. D = 4(2k-1)^2 - 4(k^2)(4) = 4(4k^2 - 4k + 1) - 16k^2 = 16k^2 - 16k + 4 - 16k^2 = -16k + 4. Setting -16k + 4 = 0 gives k = 4/16 = 1/4.

AI explanation

For the quadratic equation k^2x^2 - 2(2k - 1)x + 4 = 0 to have real and equal roots, the discriminant (b^2 - 4ac) must be zero. Substituting a = k^2, b = -2(2k - 1), and c = 4 gives [-2(2k - 1)]^2 - 4(k^2)(4) = 0. This expands to 4(4k^2 - 4k + 1) - 16k^2 = 0, which simplifies to 16k^2 - 16k + 4 - 16k^2 = 0. Combining like terms leaves -16k + 4 = 0; solving this gives 16k = 4, so k = 1/4. Thus, the statement is True.

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