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

Guest Aug 28, 2017

1+0 Answers


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, \(c=\pm\sqrt[4]{34}\)

TheXSquaredFactor  Aug 28, 2017

13 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details