simplify an expression with two square roots using quotient rule

Mar 2020
5
0
milwaukee
√(9a⁵b¹⁴/3a⁴b⁵)

I've done the next step and put the two sqrts as one radical with the radicand being a quotient.
 

romsek

Math Team
Sep 2015
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\(\displaystyle \sqrt{\dfrac{9a^5 b^{14}}{3a^4b^5}} = \\
\sqrt{\dfrac{(3a^2b^7)^2 a}{3b(a^2b^2)^2}} = \\

\dfrac{3a^2b^7}{a^2b^2}\sqrt{\dfrac{a}{3b}}
\)
 
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Mar 2020
5
0
milwaukee
well the book gives this answer b ⁴ √ 3ab as the correct solution. is the book wrong? I thought it might be incorrect as the 4th power outside the radical is not a perfect square of b
 
Mar 2020
5
0
milwaukee
\(\displaystyle \sqrt{\dfrac{9a^5 b^{14}}{3a^4b^5}} = \\
\sqrt{\dfrac{(3a^2b^7)^2 a}{3b(a^2b^2)^2}} = \\

\dfrac{3a^2b^7}{a^2b^2}\sqrt{\dfrac{a}{3b}}
\)

you see in the middle sq rt.... what is that called where you powered the numerator and denominator by 2 to free it from being the radicand?
 
Mar 2020
5
0
milwaukee
and also the book says you divide the quotient that's the radicand while the exponents for division use the quotient rule of exponents

which the sqrt of b9 is b⁴ because (b⁴)² = b⁸ with the leftover b in the radicand
 

romsek

Math Team
Sep 2015
2,954
1,666
USA
and also the book says you divide the quotient that's the radicand while the exponents for division use the quotient rule of exponents

which the sqrt of b9 is b⁴ because (b⁴)² = b⁸ with the leftover b in the radicand
jeeze you're right I should have simplified that further.

Let's even take a different approach

\(\displaystyle \sqrt{\dfrac{9a^5 b^{14}}{3a^4b^5}} = \\~\\

\sqrt{3ab^9} = \\~\\

b^4 \sqrt{3ab}

\)
 
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