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For what values of j does the equation (2x + 7)(x - 5) = -43 + jx + x^2 have exactly one real solution? Express your answer as a list of numbers, separated by commas.

 Nov 15, 2021
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  (2x + 7)(x - 5) = -43 + jx + x2
    2x2 - 3x - 35 = -43 + jx + x2
 x2 - 3x - jx + 8 = 0
x2 - (3 + j)x + 8 = 0

when there is only one real solution, by the quadratic equation:
                b2 - 4ac = 0
(-(3 + j))2 - 4(1)(8) = 0
              (-(3 + j))2 = 32
                  -(3 + j) = \(±\sqrt{32}\)
                     3 + j = \(±4\sqrt{2}\)
                           j = \(±4\sqrt{2} - 3\)


\(4\sqrt{2} - 3 ≈ 2.66\)
\(-4\sqrt{2} - 3 ≈ -8.66\)

 

which then your answer is -8.66, 2.66

please tell me if I did anything wrong. hope this helps

 Nov 15, 2021

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