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Suppose that \(a\) is a nonzero constant for which the equation \(ax^2+20x+7=0\) has only one solution. Find this solution.

 Jun 28, 2021
edited by Guest  Jun 28, 2021
 #1
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Completing the square, we get (20/7*x + 7)^2 = 0, so a = (20/7)^2 = 400/49.

 Jun 28, 2021
 #2
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The discriminant in the quadratic formula must equal 0

 

\(\triangle=0\\ b^2-4ac=0\\ 400-28a=0\\ 400=28a\\ a=14\frac{2}{7}\)

 Jun 28, 2021

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