We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

+0

# HELP!!!

+1
54
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

### 2+0 Answers

#1
+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
+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, $$\boxed{x=-\frac{2}{3}}.$$

.
Mar 17, 2019