# simplify an expression with two square roots using quotient rule

#### kenquotient

√(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
$$\displaystyle \sqrt{\dfrac{9a^5 b^{14}}{3a^4b^5}} = \\ \sqrt{\dfrac{(3a^2b^7)^2 a}{3b(a^2b^2)^2}} = \\ \dfrac{3a^2b^7}{a^2b^2}\sqrt{\dfrac{a}{3b}}$$

topsquark

#### kenquotient

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

#### kenquotient

$$\displaystyle \sqrt{\dfrac{9a^5 b^{14}}{3a^4b^5}} = \\ \sqrt{\dfrac{(3a^2b^7)^2 a}{3b(a^2b^2)^2}} = \\ \dfrac{3a^2b^7}{a^2b^2}\sqrt{\dfrac{a}{3b}}$$

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?

#### kenquotient

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
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

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

topsquark

#### kenquotient

aweee thankyou so much... you've helped me a great deal...