First, we need to expand the (2x + 2)^2(2x+2)2 term using this rule for quadratics:
(color(red)(x) + color(blue)(y))^2 = (color(red)(x) + color(blue)(y))(color(red)(x) + color(blue)(y)) = color(red)(x)^2 + 2color(red)(x)color(blue)(y) + color(blue)(y)^2(x+y)2=(x+y)(x+y)=x2+2xy+y2
(color(red)(2x) + color(blue)(2))^2 = (color(red)(2x) + color(blue)(2))(color(red)(2x) + color(blue)(2)) =(2x+2)2=(2x+2)(2x+2)=
(color(red)(2x))^2 + (2 xx color(red)(2x) xx color(blue)(2)) + color(blue)(2)^2 = 4x^2 + 8x + 4(2x)2+(2×2x×2)+22=4x2+8x+4
Substituting and simplifying gives:
y = -(2x + 2)^2 + 6x^2 + 3x - 12y=−(2x+2)2+6x2+3x−12
y = -(4x^2 + 8x + 4) + 6x^2 + 3x - 12y=−(4x2+8x+4)+6x2+3x−12
y = -4x^2 - 8x - 4 + 6x^2 + 3x - 12y=−4x2−8x−4+6x2+3x−12
y = -4x^2 + 6x^2 - 8x + 3x - 4 - 12y=−4x2+6x2−8x+3x−4−12
y = (-4 + 6)x^2 + (-8 + 3)x + (-4 - 12)y=(−4+6)x2+(−8+3)x+(−4−12)
y = 2x^2 + (-5)x + (-16)y=2x2+(−5)x+(−16)
y = 2x^2 - 5x - 16y=2x2−5x−16
We can now use the quadratic equation to solve this problem:
The quadratic formula states:
For color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0ax2+bx+c=0, the values of xx which are the solutions to the equation are given by:
x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))x=−b±√b2−(4ac)2⋅a
Substituting:
color(red)(2)2 for color(red)(a)a
color(blue)(-5)−5 for color(blue)(b)b
color(green)(-16)−16 for color(green)(c)c gives:
x = (-color(blue)(-5) +- sqrt(color(blue)(-5)^2 - (4 * color(red)(2) * color(green)(-16))))/(2 * color(red)(2))x=−−5±√−52−(4⋅2⋅−16)2⋅2
x = (5 +- sqrt(25 - (-128)))/4x=5±√25−(−128)4
x = (5 +- sqrt(25 + 128))/4x=5±√25+1284
x = (5 +- sqrt(153))/4x=5±√1534
x = (5 +- sqrt(9 xx 17))/4x=5±√9×174
x = (5 +- sqrt(9)sqrt(17))/4x=5±√9√174
x = (5 +- 3sqrt(17))/4x=5±3√174
