How do you factor and solve #x^2-7x+8=0#?

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Answer 1 · 2018-01-18T02:38:21

#x=(7+-sqrt(17))/2#

We cannot actually factor this into simple terms, but we can use a similar method, called completing the square.

#x^2-7x+8=0#

#x^2-7x=-8#

Now, we need the LHS to be in the form of #x^2+2ax+a^2=(x+a)^2#.

In this case, #2ax=-7x#

#a=-7/2#

Therefore, the equation becomes

#x^2-7x+(-7/2)^2=-8+(-7/2)^2#

Now, we can factorize the expression

#(x-7/2)^2=-8+49/4#

#(x-7/2)^2=17/4#

#x-7/2=+-sqrt(17/4)#

Simplifying the right side, we get

#x-7/2=(+-sqrt(17))/2#

Now, we can add #7/2# to both sides to get,

#x=7/2+-(sqrt(17))/2#

#x=(7+-sqrt(17))/2#