How do you find the roots for #x^2 + 24 = –11x#?

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Answer 1 · 2015-06-05T21:13:10

First add #11x# to both sides to get

#x^2+11x+24 = 0#

#24 = 3 xx 8# and #11 = 3 + 8#

So: #x^2+11x+24 = (x+3)(x+8)#

This is based on the general observation that:

#(x+a)(x+b) = x^2 + ax + bx + ab#

#= x^2 + (a+b)x + ab#

Now #(x+3)(x+8) = 0# when #x=-3# or #x=-8# so those are the roots of #x^2+11x+24 = 0#