How do you write a quadratic equation in standard form that has two solutions, 6, and -10 with the leading coefficient one?

1 Answer
Sep 29, 2016

y=x^2+4x-60

Explanation:

We want an equation which equals 0 at the given points 6 and -10.

Our quadratic equation should be a product of expressions which are zero at the specified roots.
Consider (x-6)*(x+10) = 0

This equality holds if x=6 since
(6-6)*(6+10) = 0*16 = 0
And the equality holds if x=-10 since
(-10-6)*(-10+10) = -16*0 = 0

Expanding this equation by the FOIL method, we get:

x^2 + 10x - 6x - 60

Combining like terms, we find our solution:

x^2 +4x - 60