How do you factor and solve #x^2-16=6x#?

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Answer 1 · 2017-02-11T11:42:31

#x=-2, 8#

1) Rearrange into the form#" "ax^2+bx+c=0#

2) Factorise

3) Put each bracket #=0# and solve each bracket

1) Rearrange

#x^2-16=6x#

subtract #6x# from both sides

#x^2-16-6x=6x-6x#

#x^2-16-6x=0#

#x^2-6x-16=0#

2) Factorise

#x^2-6x-16=0#

a) #"because it is an " x^2 " equation, the brackets will be of the form " (x+-" ")(x+-" ")#

we now need to find the two numbers for the brackets.

#x^2color(red)(-6)xcolor(blue)(-16)=0#

we need two numbers that multiply to #color(blue)(-16)#, but add to #color(red)(-6)#

by trial and error:

#-16xx1=-16=>-16+ 1=-15" X"#

#-8xx2=-16=>-8+ 2=--6" sqrt#

so the brackets now become;

#(x-8)(x+2)=0#

3) Solve

#x-8=0=>x=8#

#x+2=0=>x=-2#

#:.x=-2, 8#