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

 Aug 28, 2017

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.


Now, we must check that \(\pm\sqrt[4]{34}\) are valid solutions to this equation. I'll check them separately.


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


Now, let's check the other answer.


\(\sqrt{\sqrt{34}}=\sqrt{\left(-\sqrt[4]{34}\right)^2}\) This time, we cannot use the rule we used before because \(-\sqrt[4]{34}<0\). 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, \(c=\pm\sqrt[4]{34}\)

 Aug 28, 2017

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