How do you find the roots, real and imaginary, of $y = - 3 x^{2} - 16 x + {(x - 3)}^{2} + 2$ $y= -3x^2-16x +(x-3)^2+2$ using the quadratic formula?

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Answer 1 · 2017-12-02T20:55:44

$2$ $2$ real solutions:

$x\approx -11.47913\qquad,\qquad x\approx 0.47913$

First let’s simplify the equation by FOILing the $(x-3)^2$ and combining like terms:

$(x-3)^2\quad\implies \quadx^2-3x-3x+9\quad\implies \quadx^2-6x+9$

Now we can substitute that for $(x-3)^2$ in the original equation:

$color(blue)(-3x^2)-color(red)(16x)+(color(blue)(x^2)-color(red)(6x)+color(green)(9))+color(green)(2)$

Combine like terms:

$color(blue)(-3x^2)+color(blue)(x^2)-color(red)(16x)-color(red)(6x)+color(green)(9)+color(green)(2)$

$\implies color(blue)(-22x)-color(red)(22x)+color(green)(11)$

Now let’s define the variables:

And plug them into the quadratic formula:

$x=\frac{color(red)(-b)\pm\sqrt{color(red)(b^2)-4(color(blue)(a))(color(green)(c))}}{2 color(blue)(a)}$

$\implies x=\frac{color(red)(22)\pm\sqrt{(color(red)(-22))-4(color(blue)(-2))(color(green)(11))}}{2(color(blue)(-2))$

$\implies x=\frac{color(red)(22)\pm\sqrt{color(red)(484)+88}}{-4}$

$\implies x=\frac{color(red)(22)\pm\sqrt{572}}{-4}$

$\implies x=\frac{color(red)(22)\pm 2\sqrt{143}}{-4}$

$\implies x=\frac{11\pm1\sqrt{143}}{-2}$

$\implies x\approx -11.47913\qquad,\qquad x\approx 0.47913$

How do you find the roots, real and imaginary, of y= -3x^2-16x +(x-3)^2+2 y=−3x2−16x+(x−3)2+2y= -3x^2-16x +(x-3)^2+2 using the quadratic formula?