How do you factor completely #2x^2+20=x^2+9x#?

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Answer 1 · 2017-12-28T20:03:45

GIven: #2x^2+20=x^2+9x#

Combine terms so that the quadratic is equal to 0:

#x^2-9x+20=0#

This will factor into #(x-r_1)(x-r_2) = 0#, if we can find numbers such that #r_1r_2 = 20# and #-(r_1+r_2) = -9#.

4 and 5 will do it #(4)(5) = 20# and #-(4+5) = -9#:

#(x - 4)(x-5) = 0#

Answer 2 · 2017-12-28T20:08:15

Set it equal to zero, then find factors of #c# that add to #b#.

Set equal to zero:

#x^2-9x+20=0#

Looking at the discriminate: (#b^2-4ac#)

#81-4*1*20=81-80=1#

Since #1# is a perfect square, we know it factors.

Factors of #20# that add to #-9# are #-5# and #-4#.

#(x-5)(x-4)=x^2-9x+20#