+0

# HELP!!!

+1
149
2

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
+7713
+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
+4323
+3

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, $$\boxed{x=-\frac{2}{3}}.$$

.
Mar 17, 2019