SOLUTION: Find k such that the equation x^2 - kx + 4 = 0 has a repeated real solution. Let me see. I think the discriminant applies here. b^2 - 4ac = 0 k^2

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Question 1207457: Find k such that the equation x^2 - kx + 4 = 0 has a repeated real solution.

Let me see.

I think the discriminant applies here.

b^2 - 4ac = 0

k^2 - 4(1)(4) = 0

k^2 - 4(4) = 0

k^2 - 16 = 0

k^2 = 16

sqrt{k^2} = sqrt{16}

k = -4 or k = 4

A. Is this correct?
B. If so, does k = -4 or does k = 4?
Answer by ikleyn(52778) (Show Source):
You can put this solution on YOUR website!
.

(A) It is correct. (B) Both k= -4 and k= 4 give repeated root. In case k= -4, the polynomial is x^2 + 4x + 4 = with repeated real root x= -2. In case k= 4, the polynomial is x^2 - 4x + 4 = with repeated real root x= 2.

Solved and answered with complete explanation.

SOLUTION: Find k such that the equation x^2 - kx + 4 = 0 has a repeated real solution. Let me see. I think the discriminant applies here. b^2 - 4ac = 0 k^2 - 4(1)(4) =