Question

HELP PLEASE!

What is the x-coordinate of one solution to the following system?

y=6x^2+12x+2

y=4x^2+16x+16

a. -1.83

b. -11.31

c. 1.00

d. 13.67

Answer 1

a. -1.83

Explanation:

`y=6x^2+12x+2\qquad(1)\\\\y=4x^2+16x+16\qquad(2)\\\\\text{Substitute (1) to (2):}\\\\6x^2+12x+2=4x^2+16x+16\\\\\text{subtract}\ 4x^2,\ 16x\ \text{and}\ 16\ \text{from both sides}\\\\6x^2-4x^2+12x-16x+2-16=4x^2+16x+16-4x^2-16x-16\\\\\text{combine like terms}\\\\(6x^2-4x^2)+(12x-16x)+(2-16)=(4x^2-4x^2)+(16x-16x)+(16-16)\\\\2x^2-4x-14=0\\\\\text{divide both sides by 2}\\\\(2x^2)/(2)-(4x)/(2)-(14)/(2)=(0)/(2)\\\\x^2-2x-7=0`

`x^2-2x-7=0\\\\\text{add 7 to both sides}\\\\x^2-2x-7+7=0+7\\\\x^2-2x=7\\\\\text{add}\ 1^2\ \text{to both sides}\\\\x^2-2x+1^2=7+1^2\\\\\text{use}\ (a-b)^2=a^2-2ab+b^2\\\\(x-1)^2=8\iff x-1=\pm\sqrt8\\\\\text{add 1 to both sides}\\\\x-1+1=\pm\sqrt8+1\\\\x=1\pm\sqrt8\\\\\sqrt8\approx2.83\to x\approx1\pm2.83\\\\x\approx-1.83\ \vee\ x=3.83`