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

Feb 7, 2022

#1
+36416
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(2x + 7)(x + 43) = -5 + jx     expands and simplifies to

2x^2 + 86x + 7x + 301  - jx  + 5 = 0

2x^2 93x-jx + 306  = 0

If the disciminant = 0 then there is only one solution

(b^2) - 4ac = 0

(93-j)^2 - 4( 2)(306) = 0

8649 - 186j + j^2  - 2448 = 0

j^2 -186j + 6201  = 0                           Use quadratic Formula to find j = 93+- 12 sqrt(17)

Feb 7, 2022
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