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

Answer:

#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#