How do you find the roots, real and imaginary, of #y=-2(x - 2)^2-x^2+x + 12# using the quadratic formula?

1 Answer
Aug 17, 2017

Answer:

You gotta re-write it in the proper form...

Explanation:

Remember that your quadratic equation is meant to find x for equations of form

#ax^2 + bx + c = 0#

So you start by:

#y = -2(x - 2)^2 - x^2 + x + 12#

#= -2(x^2 - 4x + 4) - x^2 + x + 12#

#= -2x^2 + 8x -8 - x^2 + x + 12#

#= -3x^2 + 9x + 4#

...so now you have your coefficients a, b, and c for the quadratic equation.

#a = -3, b = 9, c = 4#

Plug these in to the quadratic formula:

#x = (-b +- sqrt(b^2 - 4ac))/(2a)#

I get roughly #x = 3.393 and -0.393#

It's always good to check your answers by plugging these values for x into your original equation. You should get zero. (which you don't, with these values, exactly, because I've rounded off. But it's very close.)

GOOD LUCK