How do you find the roots, real and imaginary, of $y=-5(x-3)^2-45$ using the quadratic formula?

Question

2110 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-04T19:54:04

$x=-3/2+-9/2i$

To make this more strait forward lets convert the equation back to the form of $y=ax^2+bx+c$

Square the brackets.

$y=-5(x^2-3x+9)-45$

Multiply out the -5

$y=-5x^2+15x-45-45$

$y=-5x^2+15x-90$

Thus the quadratic formula states that

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

where $a=-5"; "b=+15"; "c=-90$

$x=(-15+-sqrt((-15)^2-4(-5)(-90)))/(2(-5))$

$x=-3/2+-(sqrt(-2025))/(-10)$

If you are ever not sure about factors do a quick sketch of a prime factor tree on the side of the page.
Tony B

$x=-3/2+-(sqrt(5^2xx9^2xx(-1)))/(-10)$

$x=-3/2+-( cancel(5)^1xx9xxi)/(cancel(10)^2)$

As there is a $+-$ sign in front of $(9xxi)/2$ you can ignore any negative within it.

$x=-3/2+-9/2i$

How do you find the roots, real and imaginary, of y=-5(x-3)^2-45 y=-5(x-3)^2-45 using the quadratic formula?