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29 December, 07:34

For what values of j does the equation (2x+7) (x-5) = - 43+jx have exactly one solution

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  1. 29 December, 07:58
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    Expand and rearrange the terms:

    2x²-10x+7x-35=-43+jx

    2x² - (3+j) x+8=0

    x²-[ (3+j) / 2]x+4=0

    for the equation to have exactly one solution, it has to be a square. Recall that (a-b) ²=a²-2ab+b². In this case, a=x, b=2, 2ab=2*x*2=4x

    so (3+j) / 2=4

    3+j=8

    j=5
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