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# I need help fast!

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I need to find the square root answer of Square root of 34 = square root of C squared

Aug 28, 2017

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Is the equation $$\sqrt{\sqrt{34}}=\sqrt{c^2}$$ in which you want solved? This is what I intepret it as. I will solve it:

 $$\sqrt{\sqrt{34}}=\sqrt{c^2}$$ Square both sides of the equation to get rid of one set of square roots. $$\sqrt{34}=c^2$$ Take the square root of both sides. Of course, taking the square root results in a positive and negative answer. $$c=\pm\sqrt{\sqrt{34}}$$ Now, let's simplify.  $$\sqrt{\sqrt{34}}$$Note that $$\sqrt[n]{a}=a^{\frac{1}{n}}$$. $$\sqrt{\sqrt{34}}=(\sqrt{34})^{\frac{1}{2}}$$ Let's use the same principle as above. $$(\sqrt{34})^{\frac{1}{2}}=\left(34^{\frac{1}{2}}\right)^{\frac{1}{2}}$$ In the current situation we are in, we will use the rule that $$\left(a^b\right)^c=a^{b*c}$$. $$\left(34^{\frac{1}{2}}\right)^{\frac{1}{2}}=34^{\frac{1}{2}*\frac{1}{2}}=34^{\frac{1}{4}}$$ Yet again, we will use the inverse of $$\sqrt[n]{a}=a^{\frac{1}{n}}$$ to simplify. $$c=\pm\sqrt{34}\approx\pm2.4147$$

Now, we must check that $$\pm\sqrt{34}$$ are valid solutions to this equation. I'll check them separately.

 $$\sqrt{\sqrt{34}}=\sqrt{c^2}$$ Plug in the value of $$\sqrt{34}$$ and check its validity. $$\sqrt{\sqrt{34}}=\sqrt{\left(\sqrt{34}\right)^2}$$ This may look complicated, but it can be simplified by realizing that$$\sqrt{a^2}=a$$, assuming $$a\geq0$$ $$\sqrt{\sqrt{34}}=\sqrt{34}$$ We already calculated earlier than the square root of the square root of a number has a square root index of 4. $$\sqrt{34}=\sqrt{34}$$

Now, let's check the other answer.

 $$\sqrt{\sqrt{34}}=\sqrt{\left(-\sqrt{34}\right)^2}$$ This time, we cannot use the rule we used before because $$-\sqrt{34}<0$$. Let's rewrite this as an exponent. $$\sqrt{\sqrt{34}}=\sqrt{\left(-34^{\frac{1}{4}}\right)^2}$$ Let's use the exponent rule we have used before that states that $$\left(a^b\right)^c=a^{b*c}$$. Note that squaring a number automatically makes a number positive, so that is why the negative sign disappears in the next step. $$\left(-34^{\frac{1}{4}}\right)^2=34^{\frac{1}{4}*\frac{2}{1}}=34^{\frac{2}{4}}=34^{\frac{1}{2}}=\sqrt{34}$$ Reinsert this into the equation and compare. $$\sqrt{\sqrt{34}}=\sqrt{\sqrt{34}}$$ This statement is true, as both sides are the same, so both values for c are solutions.

Therefore, $$c=\pm\sqrt{34}$$

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Aug 28, 2017