How do you write #x^2-8x+13=0# in the form #(x-a)^2=b#, where #a# and #b# are integers?

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Answer 1 · 2017-07-16T17:03:04

#(x-4)^2=3#

We need to 'force' it to give us the target format.

If we start with #(x-4)(x-4)# this gives us: #x^2-8x+16#

So we have now achieved the correct value of #-8x# but the constant is wrong.

So we now need to change the 16 to give us the constant of 13. This can be achieved by subtracting 3

#(x-4)(x-4)-3#

But the right hand side is 0 so we write:

#(x-4)(x-4)-3=0#

Add 3 to both sides

#(x-4)(x-4)=3#

#(x-4)^2=3# matching the target format