2
$\begingroup$

Question:

Simplify

$$ \left(ab \sqrt[4]{a^{3}/\sqrt{b\sqrt{b}}}\right)^{2}$$

Attempted solution:

Rewriting it to look a bit better:

$$\left(ab \sqrt[4]{\frac{a^{3}}{\sqrt{b\sqrt{b}}}}\right)^{2}$$

Rewriting the denominator by replacing square roots with powers of fractions:

$$\left(ab \sqrt[4]{\frac{a^{3}}{(b (b^{\frac{1}{2}}))^{\frac{1}{2}}}}\right)^{2}$$

Combining b:s in the denominator:

$$\left(ab \sqrt[4]{\frac{a^{3}}{b^{\frac{3}{4}}}}\right)^{2}$$

Distributing the 4th root:

$$\left(\frac{ab \cdot a^{\frac{12}{16}}}{b^{\frac{3}{16}}}\right)^{2}$$

Combining a:s with a:s and b:s with b:s:

$$\left(b^{\frac{13}{16}} a^{\frac{28}{16}}\right)^{2}$$

Squaring gives the final result:

$$b^{26} a^{56}$$

The answer turns out to be:

$$a^{\frac{7}{2}} b^{\frac{13}{8}}$$

This is quite far away from the result I reached. Where did I go wrong, and are there are key insights that are useful for solving questions with lots and lots of square, cube and higher roots?

$\endgroup$
0
5
$\begingroup$

Note that $$\left(b^{\frac{13}{16}}a^{\frac{28}{16}}\right)^2=b^{\frac{13}{16}\cdot 2}\times a^{\frac{28}{16}\cdot 2}.$$

$\endgroup$
2
$\begingroup$

$\left(b^{13\over16}a^{28\over16}\right)^2=b^{26\over16}a^{56\over16}=b^{13\over8}a^{7\over2}$

Because $\left(x^y\right)^z=x^{yz}$

$\endgroup$
1
$\begingroup$

we have $a^2b^2\sqrt{\frac{a^3}{b^{3/4}}}=\sqrt{\frac{a^4b^4a^3}{b^{3/4}}}=a^{7/2}(b^{13/4})^{1/2}=a^{7/2}b^{13/8}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.