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,

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

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

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