Need help can someone explain how to do it?

Question

Perform the indicated operation and simplify the radical.

7√2 • 4√8 = ?

Perform the indicated operation and simplify the radical.

√16a³b
‾‾‾‾‾‾‾‾‾‾‾=?
√24a²b²

Answer 1

You can multiply the numbers inside and outside the radical separately:

(7V2) * (4V8) = (7*4) V(2*8) = 28V16

and V16 can be simplified as 4, so 28 * 4 = 112.



Second problem is similar: divide the expressions inside the radical.

Answer 2

7√2 • 4√8 = 7√2 • 4√(4*2) = 7√2 • 8√2 = 56√2 • √2 = 56*2 =112

You can see that basically, you find any squares you can under the radical then take them out, for example breaking the 8 into 4 times two allows you to take the square root of 4, which is two and multiply the number in front of the radical. Similarly in the next example (I assume only the numbers are under the radical), the 16 is 4 squared so take it out of the radical, and 24 is equal to 4*6 so we can take the square root of four out.

√16a³b
‾‾‾‾‾‾‾‾‾‾‾=
√24a²b²

2*a
‾‾‾‾‾
√6*b

Answer 3

For the top one, because they are all multiplications you can simply multiply them all together. If you need to multiply together two sqrts, you can take the square root of the multiplication of the numbers.

So, sqrt (A)*sqrt(B)=sqrt(A*B).
So, sqrt(3)*sqrt(4)=sqrt(12)
Or sqrt(4)*sqrt(16)=sqrt(64)=8 (you can confirm it with this one by doing the sqrts first and then multiplying:- sqrt(4)=2, sqrt(16)=4, 2*4=8).

So 7*sqrt(2)*4*sqrt(8)=7*4 * sqrt(2)*sqrt(8). You should be able to do this now hopefully.


For the second one, I assume you understand about canceling out powers above and below equations.

If you have a sqrt above and a sqrt below division, just like multiplication. you can divide inside the sqrt sign, so:
sqrt(A) = sqrt( A/B)
--------
sqrt(B)

Again you can test it out:
sqrt (64)
----------= sqrt(64/16)=sqrt(4)=2 (=8/4 )
sqrt(16)

So in the second one, group together the square roots first and then cancel out the powers of the a's and b's.

Good luck.

Answer 4

4 sqrt(8) = 8(sqrt(2))

so 56*sqrt(2)*sqrt(2) = 112

16a^3b/24a^2b^2 = 2a/3b

Answer 5

I think the second answer should be:

2√a
‾‾‾‾‾‾‾‾‾‾‾
√(6b)

Answer 6

7sqrt(2) * 4sqrt(8)
7sqrt(2) * 4sqrt(4 * 2)
7sqrt(2) * 8sqrt(2)
(7 * 8) * (sqrt(2))^2
56 * 2
112

-------------------------------------

(sqrt(16a^3b))/(sqrt(24a^2b^2))
(sqrt(16) * sqrt(a^3) * sqrt(b))/(sqrt(24) * sqrt(a^2) * sqrt(b^2))
(4sqrt(a^2 * a) * sqrt(b))/(sqrt(4 * 6) * a * b)
((4a)sqrt(ab))/((2ab)sqrt(6))

((4a)/(2ab)) * (sqrt(ab)/sqrt(6))
(2/b) * (sqrt(6ab))/6

(2/6) * (sqrt(6ab))/b)
(1/3) * (sqrt(6ab))/b

ANS : (sqrt(6ab))/(3b)

or you can do it like this

(sqrt(16a^3b))/(sqrt(24a^2b^2))
sqrt((16a^3b)/(24a^2b^2))
sqrt((16/24) * ((a^3)/(a^2)) * (b/(b^2)))
sqrt((4/6) * a^(3 - 2) * b^(1 - 2))
2sqrt((1/6) * a * b^-1)
2sqrt((1/6) * a * (1/b))
2sqrt(a/(6b))
(2sqrt(a))/(sqrt(6b))
(2sqrt(6ab))/(6b)
(2/6) * (sqrt(6ab))/b
(1/3) * (sqrt(6ab))/b
(sqrt(6ab))/(3b)