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.
+130561 9x^2 -18x - 16 = 0
15x^2 + 28x + 12 = 0
Since both quadratics = 0, set them = to each other
9x^2 -18x - 16 = 15x^2 + 28x + 12 subtract 9x^2 from both sides, add 18x, 16 to both sides
6x^2 + 46x + 28 = 0 divide everything by 2
3x^2 + 23x + 14 = 0 factor
(3x + 2) (x + 7) = 0 setting both factors to 0 and solving, x = -2/3 or x = -7
However since we're trying to find where both functions = 0........the solution is x = -2/3
The second solution of x = -7 is another intersection point of both graphs
Solve for x:
15 x^2+28 x+12 = 9 x^2-18 x-16
Subtract 9 x^2-18 x-16 from both sides:
6 x^2+46 x+28 = 0
The left hand side factors into a product with three terms:
2 (x+7) (3 x+2) = 0
Divide both sides by 2:
(x+7) (3 x+2) = 0
Split into two equations:
x+7 = 0 or 3 x+2 = 0
Subtract 7 from both sides:
x = -7 or 3 x+2 = 0
Subtract 2 from both sides:
x = -7 or 3 x = -2
Divide both sides by 3:
Answer: | x = -7 or x = -2/3
