Question #e1353

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Question #e1353

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Answer 1 · 2016-11-19T22:03:53

$x=\pm\sqrt(14)$

Explanation:

Equation given

$x^2-2=12$

Method 1

Isolate the $x^2$ and solve by square root.
$x^2\cancel(-2)\cancel(\color(blue)(+2))=12\color(blue)(+2)$
$x^2=14$
$x=\pm\sqrt(14)$

Method 2

Form a quadratic equation by making one side equal to 0, then solve by factoring said quadratic.
$x^2-2\color(red)(-12)=\cancel(12)\cancel(\color(red)(-12))$
$x^2-14=0$ $lArr$ the factor of variable $x$ is 0x, so...
$\rArrx^2+0x-14=0$ apply quadratic formula

quadratic formula below
$x=(-0\pm\sqrt(0^2-4(1)(-14)))/(2(1))=(\pm\sqrt(56))/2=(\pm2\sqrt(14))/2$
$\rArr(\pm\cancel(2)\sqrt(14))/\cancel(2)=\pm\sqrt(14)$

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Question #e1353

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Explanation:

Equation given

Method 1

Method 2

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