How do you solve #b^ { 2} - 9b = - 8#?

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Answer 1 · 2018-03-03T18:01:09

#b^2 -9b=-8#

or, #b^2 -9b+8=0#

or, #b^2 -8b -b +8=0#

or, #b(b-8) -1(b-8) =0#

so, #(b-8) (b-1) =0#

so, #b-8 =0# or, #b=8#

and, #b-1=0# or, #b=1#

Answer 2 · 2018-03-03T18:07:38

#b=1 or b=8#

#b^2 - 9b =-8#

Start by subtracting #-8# from both sides

#b^2-9b-(-8)=-8-(-8)#

#b^2-9b+8=-8+8#

#b^2-9b+8=0#

Now we need to factorize the left side

#(b-1)(b-8)=0#

Set factors equal to #0#

#b-1=0 or b-8=0#

#b = 0 + 1 or b = 0 +8#

#b = 1 or b= 8#