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Determine all real numbers a such that the inequality \(|x^2 + 2ax + 3a|\le2\) has exactly one solution in x.

 Dec 31, 2019
 #1
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Complete the square

\(f(x) = (x^2+2ax+a^2) + (3a-a^2) = (x+a)^2 + (3a-a^2)\)

So the minimum value is 3a-a^2

so \(3a - a^2 = 2\)

a=1,2

 Dec 31, 2019

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