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Let \(x\) be a value such that \(9x^2 - 18x - 16 = 0\) and \(15x^2 + 28x + 12 = 0.\) What is the value of \(x\)? Express your answer as a simplified common fraction.

 Mar 17, 2019
 #1
avatar+7424 
+2

\(9x^2-18x-16=0\\ (3x-8)(3x+2) = 0\\ x = \dfrac{8}{3}\text{ OR }x = \dfrac{-2}{3}\)

 

\(15x^2+28x+12 = 0\\ (3x+2)(5x+6) = 0\\ x = \dfrac{-2}{3} \text{ OR } x = \dfrac{-6}{5}\)

 

Only x = -2/3 satisfies both equations. Therefore the answer is -2/3.

 Mar 17, 2019
 #2
avatar+4077 
+2

Since they both are equal to 0, we can set them equal to each other.

Therefore, \(9x^2-18x-16=15x^2+28x+12\), and by solving the quadratic, we get two solutions, \(x=-7,\:x=-\frac{2}{3}\), but by guess and check, \(\boxed{x=-\frac{2}{3}}.\)

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 Mar 17, 2019

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