How do you solve ${(x - 9)}^{2} = 12$ $(x-9)^2=12$ ?

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How do you solve ${(x - 9)}^{2} = 12$ $(x-9)^2=12$ ?

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Answer 1 · 2018-03-30T02:39:47

$x = - 2 \sqrt{3} + 9, x = - 2 \sqrt{3} + 9$ $x=-2sqrt3+9, x=-2sqrt3+9$

Explanation:

You could expand the left side, move everything to the left, and apply the quadratic formula; however, taking the root of both sides is far faster:

$\sqrt{{(x - 9)}^{2}} = \pm \sqrt{12}$ $sqrt((x-9)^2)=+-sqrt(12)$

Recall that $\sqrt{a^{2}} = a$ $sqrt(a^2)=a$ , and that $\sqrt{12} = \sqrt{4 \cdot 3} = \sqrt{4} \sqrt{3} = 2 \sqrt{3}$ $sqrt12=sqrt(4*3)=sqrt4sqrt3=2sqrt3$ :

$x - 9 = \pm 2 \sqrt{3}$ $x-9=+-2sqrt3$

$x = - 2 \sqrt{3} + 9, x = - 2 \sqrt{3} + 9$ $x=-2sqrt3+9, x=-2sqrt3+9$

You would have the same result even with applying the quadratic formula -- this method is just faster and cleaner.

Answer 2 · 2018-03-30T02:44:14

$x = 2 \sqrt{3} + 9,$ $x=2sqrt3+9,$ $- 2 \sqrt{3} + 9$ $-2sqrt3+9$

Explanation:

Solve:

${(x - 9)}^{2} = 12$ $(x-9)^2=12$

Take the square root of both sides.

$x - 9 = \pm \sqrt{12}$ $x-9=+-sqrt12$

Prime factorize $12$ $12$ .

$x - 9 = \pm \sqrt{2 \times 2 \times 3} =$ $x-9=+-sqrt(2xx2xx3)=$

$x - 9 = \pm \sqrt{2^{2} \times 3}$ $x-9=+-sqrt(2^2xx3)$

Apply rule: $\sqrt{a^{2}} = a$ $sqrt(a^2)=a$

$x - 9 = \pm 2 \sqrt{3}$ $x-9=+-2sqrt3$

Add $9$ $9$ to both sides.

$x = \pm 2 \sqrt{3} + 9$ $x=+-2sqrt3+9$

Solutions for $x$ $x$ .

$x = 2 \sqrt{3} + 9,$ $x=2sqrt3+9,$ $- 2 \sqrt{3} + 9$ $-2sqrt3+9$

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How do you solve ${(x - 9)}^{2} = 12$ $(x-9)^2=12$ ?

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How do you solve ${(x - 9)}^{2} = 12$ $(x-9)^2=12$ ?