How do you factor #(m-7)= -(m-7)^2 + 3#?

1 Answer
George C.
Oct 29, 2016

#m = 13/2+-sqrt(13)/2#

or

#(m - 13/2 - sqrt(13)/2)(m - 13/2 + sqrt(13)/2) = 0#

Explanation:

Given:

#(m-7) = -(m-7)^2+3#

First add #(m-7)^2-3# to both sides to get:

#(m-7)^2+(m-7)-3 = 0#

This is in the form:

#ax^2+bx+c = 0#

with #x = (m-7)#, #a=color(purple)(1)#, #b=color(blue)(1)# and #c = color(brown)(-3)#

So we can use the quadratic formula to find:

#m-7 = (-b+-sqrt(b^2-4ac))/(2a)#

#color(white)(m-7) = (-color(blue)(1)+-sqrt(color(blue)(1)^2-4(color(purple)(1))(color(brown)(-3))))/(2(color(purple)(1)))#

#color(white)(m-7) = (-1+-sqrt(1+12))/2#

#color(white)(m-7) = -1/2+-sqrt(13)/2#

Add #7# to both ends to find:

#m = 13/2+-sqrt(13)/2#

Related questions
See all questions in Factoring Completely
Impact of this question
906 views around the world
You can reuse this answer
Creative Commons License