Question

(16a)/(sqrt 6a^3)

The answer key says (8 sqrt 6a)/(3a), but I can't see how.

Answer 1

\[\frac{ 16a }{ \sqrt{6a^3} }\]
This is the equation?

Answer 2

Yes

Answer 3

@MrsDavidRandle Can you help me?

Answer 4

I don't think so, sorry.

Answer 5

You have to use the rule for dividing exponents. But division is just like multiplication, except it's with reciprocal.

So check this out. You know that x^a * x^b = x^(a+b), yeah?

If you don't, then plug in a number into x, a, and b. For example 3, 4, and 5.
3^4 * 3^5 = 3*3*3*3 * 3*3*3*3*3.
But there's eight 3's in there, so that equals 3^8.

Anyway, what if b is negative? x^a * x^(-b) = x^a / x^b.

You divide them because x^(-1) = 1/x. It's the reciprocal.

So since you add exponents when you multiply, you subtract them when you divide.

Answer 6

Now, back to your problem.
\[\frac{16a}{\sqrt{6a^3}}\]
Remember that the sqrt is actually an exponent of 1/2.

Also (x^a)^b = x^(a*b).

So the denominator is actually sqrt(6) * a^(3/2), since (a^3)^(1/2) = a^(3/2), right?

Now, we're dividing a / a^(3/2). The numerator a has an exponent of 1, so 1-3/2 = -1/2.

So, a / a^(3/2) = a^(-1/2) = 1 / √a.
But it needs to be "rationalized," so that's actually √a / a.

If we rationalize the 1 / √6, we get √6 / 6.

Let's rewrite what we have so far:

\[\frac{16 \sqrt{6} \sqrt{a} }{ 6 a}\]
Now we simply divide 16/6 to get 8/3.

Answer 7

Multiplying numerator and denominator by \[\sqrt{6a ^{3}}\]
gives
\[\frac{16a \sqrt{6a ^{3}}}{6a ^{3}}\]
Dividing numerator and denominator by 2a^2 gives
\[\frac{\frac{8a}{a ^{2}}\sqrt{6a ^{3}}}{3a}\]
Now take a^2 outside the radical sign, making it a:
\[\frac{\frac{8a ^{2}}{a ^{2}}\sqrt{6a}}{3a}\]
Cancelling the a^2 gives the answer.

Answer 8

Thanks TurtleMuffin and Kropot! Kropot, your method seems kind of different from the one my textbook shows my, but it works.

Answer 9

You're welcome :)

Answer 10

Oh right, we could have rationalized from the get-go.

The beauty of math is that one can manipulate any expression in any kind of way, as long as you follow algebraic rules!