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

1 Answer
Dec 2, 2017

Answer:

#2# real solutions:

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

Explanation:

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:

  • #color(blue)(a=-2)#
  • #color(red)(b=-22#
  • #color(green)(c=11#

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#