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

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Mathematics

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17 February, 14:19

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

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Jaslene Zhang · Answer 1

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