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# Algebra

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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

#1
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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 = $$±4\sqrt{2} - 3$$

$$4\sqrt{2} - 3 ≈ 2.66$$
$$-4\sqrt{2} - 3 ≈ -8.66$$