How do you solve #x^ { 2} - 6x = 7#?

2 Answers
Karthik N
May 2, 2018

#x = 7 or x = (-1)#

Explanation:

Subtract 7 from both sides.
#implies x^2 -6x - 7 =0#
We need to factorize this expression by splitting its middle term.

#impliesx^2 -7x +x - 7= 0#
#implies x(x-7)+1(x-7) = 0#
#implies (x-7)(x+1) = 0#

#implies x - 7 = 0 or x + 1= 0#
#implies x = 7 or x = (-1)#

MAXIMILIAN C.
May 2, 2018

#x=7#
#x=-1#

Explanation:

#x^2-6x=7# Subtract #7# from both sides:
#x^2-6x-7=0#
#-7,1# are factors of #-7# that add up to #6#, so they will be factors:
#(x-7)(x+1)=0# Using the #0# product property:
#x-7=0#
#x=7#
and
#x+1=0#
#x=-1#

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