David D. answered • 06/25/18

Tutor

New to Wyzant
Former Community College Professor misses Teaching and Tutoring

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\sqrt{8^3y^5}\sqrt{2^8y}

I was trying to get a math word processor to show symbols but I am new to the Wyzant environment. But to your problem. I hope I read it correctly. You want to put the division inside one big square root, simplify (cancellations), then take the square root last of a the very simplified radicand. So make a few changes along the way. Since 8 is the third power of 2 then 8^3 is (2^3)^3 = 2^9 which is nice because you have a power (8) of 2 to cancel with, so the numerical term is 2^9/2^8 = just 2. The variable piece is y^5/y = y^4 (I am using the laws of exponents which I can review for you if you need to). Back to the radical....after the reductions of powers inside the sqrt we are left only with 2y^4, now y^(even) is always a perfect square, we can reduce the y term by taking its square root which is y^2. You can think of the sqrt as cutting exponents in half, that's why you need even powers to get away with it. The 2 is stuck inside the radical b/c it cannot be reduced. So "FINAL ANSWER" sqrt(2)y^2 (or if you prefer y^2sqrt(2). I like the first one better but technically they are both correct (commutative rule).

I hope this helped.

Dave D.