How do you solve $x^{2} - 6 x + 9 = 5$ $x^2-6x+9=5$ using square root property?

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Answer 1 · 2015-02-05T21:09:25

You need to factor $x^{2} - 6 x + 9$ $x^2-6x+9$

This will give you:

$(x - 3) (x - 3) = 5$ $(x-3)(x-3)=5$

Which of course is

${(x - 3)}^{2} = 5$ $(x-3)^2=5$

Now take the square of both sides:

$\sqrt{{(x - 3)}^{2}} = \pm \sqrt{5}$ $sqrt((x-3)^2)=+-sqrt5$

$(x - 3) = \pm \sqrt{5}$ $(x-3)=+-sqrt5$

Now solve for x

$x = 3 \pm \sqrt{5}$ $x=3+-sqrt5$

How do you solve x^2-6x+9=5x2−6x+9=5x^2-6x+9=5 using square root property?