Multiple choice

The value of k for which the roots are real and equal of the following equation (k + 1)$x^2$ + 2(k + 3)x + (k + 8) = 0 is $k = \dfrac{1}{3}$

  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, 4(k+3)^2 - 4(k+1)(k+8) = 0. (k^2 + 6k + 9) - (k^2 + 9k + 8) = 0. -3k + 1 = 0, so k = 1/3.

AI explanation

For the quadratic equation (k + 1)x^2 + 2(k + 3)x + (k + 8) = 0 to have real and equal roots, the discriminant (b^2 - 4ac) must equal zero. Substituting the coefficients gives [2(k + 3)]^2 - 4(k + 1)(k + 8) = 0. Expanding both sides yields 4(k^2 + 6k + 9) - 4(k^2 + 9k + 8) = 0; dividing the entire equation by 4 gives k^2 + 6k + 9 - k^2 - 9k - 8 = 0. Combining the terms results in -3k + 1 = 0, which means 3k = 1. Solving this gives k = 1/3, making the statement True.

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