Question

How do you find the roots, real and imaginary, of `y=-(x+5)^2+2x^2 + 5x - 12` using the quadratic formula?

Answer 1

`x_1~~6.437+0i`
`x_2~~-1.437+0i`

Explanation:

Quadratic formula, given by:
`(-b+-sqrt(b^2-4ac))/(2a)`

Before we proceed, let's us expand the equation of `y`
`y=-(x+5)^2+2x^2+5x-12`
`y=-(x^2+10x+25)+2x^2+5x-12`
`y=-x^2-10x-25+2x^2+5x-12`
`y=(2-1)x^2+(5-10)x+(-37)`
`y=(1)x^2+(-5)x+(-37)`

Determine the `a`,`b`, and `c`
`a=1`
`b=-5`
`c=-37`

Use the formula:
`x=(-b+-sqrt(b^2-4ac))/(2a)`
`x=(-(-5)+-sqrt((-5)^2-4(1)(-37)))/(2(1))`
`x=(5+-sqrt(25+37))/2`
`x=(5+-sqrt(62))/2`

`x_1=(5+sqrt(62))/2~~6.437`
`x_2=(5-sqrt(62))/2~~-1.437`

In this case, the roots are real, their imaginary parts are 0
`x_1~~6.437+0i`
`x_2~~-1.437+0i`