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By intuition it can be determined that $-7$ as well as $+7$ satisfy the condition $x^2=49$ but when I take square root both sides I get the equation $\sqrt{x^2}=\sqrt{49}$ which can only yield $+7$ due to the underoot.How do I solve algebraically for a $-7$ and $+7$ answer.

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    $\begingroup$ Difference of squares. $\endgroup$ – Karl Apr 8 '17 at 6:38
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    $\begingroup$ $\sqrt{x^2}\ne x $. $\sqrt{x^2} = |x|$. So taking square roots of both sides results in $|x| =7$ which has two solutions $x=7$ or $x=-7$. We generalize this result by stating $x^2 = k \implies x =\pm \sqrt {k} $. $\endgroup$ – fleablood Apr 8 '17 at 6:42
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$$x^2-49=(x+7)(x-7)=0$$ (difference of two squares) and one of the factors must be equal to zero. Alternatively just square root with $x=\pm 7$ (as in the quadratic formula)

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$$x^2=49$$ $$ \begin{cases} \ y=x^2 \\ y=49 \end{cases}$$ enter image description here

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$x^2= 49\not \implies x=7$

$x^2=49\implies \sqrt {x^2} =7 \not \implies x=7$ because $\sqrt{x^2}\ne x$.

$\sqrt {x^2}=|x|$ which only equals $x $ if $x$ is nonnegative. If $x $ is negative then $|x|=-x $ (which is not a negative number. $-x $ is a positive number if $x $ is negative).

So $\sqrt {x^2}=49\implies |x|=7 \implies \pm x = 7 \implies x =\pm 7$.

Both roots accounted for.

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After taking root both the side you must get the equation:$$\sqrt {x^2}=\pm\sqrt{49}$$ $$x=\pm7$$

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Yes, it does yield only $+7$. However, your LHS is not $x$, it is $|x|$, as $\sqrt{x^2}=|x|$. This is because it must be the positive root by definition. So,

$$\begin{align}x^2 &= 49\\\sqrt{x^2}&=\sqrt{49}\\|x|&=7\\ x&=\pm7\end{align}$$

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