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Let $f(x) = x^2 + 4x - 31$. For what value of $a$ is there exactly one real value of $x$ such that $f(x) = a$?

 Mar 1, 2020
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If f(x) = a then we have:  x2 + 4x -31 = a

 

Rewrite as x2 + 4x - 31 - a =0

 

For this to have a single root we must have  42 - 4*(-31 - a) = 0 or a = -35

 Mar 1, 2020
edited by Alan  Mar 1, 2020

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