Simplify Sqaure Roots?

Question

Assuming all variables are positive... how do I simplify this problem?

sqrtx^4y^3


How do I simplify this problem??

(8sqrt5) - (sqrt20) Answering in radical form ?


Also answering in radical form... simplify

sqrt20
--------
5

(minus)

1
--
sqrt5


THANK YOU !!!!

Is the answer to (sqrtx^4y^3)

x^2ysqrty or x^2sqrt(y)^3

Answer 1

#1)______________________________________

√(x^4x^3)

First, seperate into two "Square Roots".

√(x^4) √(x^3)

The root of a "Square Root" is the #2.

So to ELIMINATE the ²√ , we need to raise the INSIDE to the "Power" of 2 . . . ( )².

When the Root and the Power are the SAME. . . the INSIDE (known as The Descriminant) comes out.

But the Root and the Power MUST BE THE SAME!

Whatever is leftover is the REMAINDER under the "Square Root".

²√(x²)² ²√(x¹)² ²√(x¹)
(x²)(x¹)√(x¹)

When Multiplying w/ "like bases" (variables). . . ADD the exponents together.

(x ² + ¹)√(x)

ANSWER #1=================================

√(x^4x^3) = (x³)√(x)

==============================================

#2)_______________________________________________

[8√(5)] - √(20)

[8√(5)] - [√(4)√(5)]

[8√(5)] - [2√(5)]

ANSWER #2 =====================================

[8√(5)] - √(20) = 6√(5)

==============================================

#3)______________________________________________

{ [ √(20) ] / 5 } - [ 1/√(5) ]

{ [ √(20) ] / 5 } - { [1][√(5)] / [√(5)][√(5)] }

{ [ √(20) ] / 5 } - { [√(5)] / 5 }

{ [√(4)√(5)] / 5 } - { [√(5)] / 5 }

{ [2√(5)] / 5 } - { [1√(5)] / 5 }

{ [2√(5)] - [1√(5)] } / 5

[1√(5)] / 5

ANSWER #3 =================================

{ [ √(20) ] / 5 } - [ 1/√(5) ] = [√(5)] / 5

===========================================

Good Job making it to the end with me!
Great questions!

Good luck on your Finals! ♥

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Additional Details

9 minutes ago

The answer to (sqrtx^4y^3) is:

x^2ysqrty or as I like to write it:

√(x^4y^3) = x²y√(y)

GOOD JOB! ♥
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

Answer 2

sqrtx^4y^3: sqrt is the same as (1/2) so you can multiply the exponents (1/2*4) and you get x^2y^3

(8sqrt5)-(sqrt20): square roots of large numbers can be broken down into multiples of those numbers. sqrt20 can be broken down into sqrt5*sqrt4 sqrt4=2 and the problem becomes (8sqrt5)-(2sqrt5)=6sqrt5

sqrt20/5-1/sqrt5: multiply the 1/sqrt5 by sqrt5 to get 1sqrt5/5 you now have a common denominator of 5. replace the sqrt20 with 2sqrt5(see above) and the equation becomes 2sqrt5/5-1sqrt5/5=1sqrt5/5

Answer 3

okay, lets see.. all you do is remember that "sqrt" is the same as "to the exponent 1/2) so the first one..

(x^4*y^3)^(1/2) then use your law of exponents (multiplying one)..
= x^2*y^(3/2) = x^2*sqrt(y^3)

next, is 8sqrt(5) - sqrt(20) now, the only way you can add/subract terms with radicals if they have the same radical. knowing this.. sqrt(20) can be broken up into sqrt(4)*sqrt(5) which = 2sqrt(5)

so 8sqrt(5) - 2sqrt(5) = 6sqrt(5)

last question! again, you can simplify sqrt(20) into 2sqrt(5) so..

2sqrt(5)/(5) - 1/(sqrt(5))... to add/subtract fractions, you need a common denominator.. so we multiply the second term by sqrt(5)/sqrt(5) [which is the same as 1, so we're not changing it..] and we get..

2sqrt(5)/(5) - sqrt(5)/(5)
= sqrt(5)/(5) = (5)^(-1/2) = 1/(sqrt(5))
but generally, we keep the "demoninator rationalized" so the it's sqrt(5)/(5)

working with squareroots, or any radicals for that matter is basically just working with exponents. Just make sure you study those law of exponents and you'll be fine!

Answer 4

If everything is multiplied or divided together under the radical with no addition or subtraction, whatever you collect in squares can be taken out. For example x^7 = (x³)(x³)x so x³ can be taken out of the radical and x is left behind under the radical.

√(x^4y^3) = √{(x²)²(y²)y} = x²y√y

(√20)/5 - 1/√5 = (2√5)/5 - (√5)/5 = (√5)/5

Answer 5

1/ sqrt(x^4y^3) = (x^2)(y)(sqrt y)

2/ (8sqrt5) - (sqrt20) = (8sqrt5) - {sqrt(4x5)} = (8sqrt5) - (2sqrt5) = 6sqrt5

3/ {(sqrt20)/5} - {1/(sqrt5)

first simplify (sqrt20)/5 = {sqrt (4x5)}/5 = {2(sqrt5)}/5 = 2/ (sqrt5)

Now take the one you simplify minus the second part because they have the same common denominator now.

{2/(sqrt5)} - {1/(sqrt5)} = 1/(sqrt5) = (sqrt5)/5

never leave the square root as the answer for the denominator.

Answer 6

one way is to right component the quantity, and write out the variable, and then for each pair of equivalent aspects, positioned one among them outdoors the ? (and remove the different). Like for yours, you'll have ?3•3•3•a•a one pair of three's and one pair of a's, with a three left over so 3a?3 also, for a classic sq. root, you'll locate the biggest perfect sq. component of it, write the quantity as a multiply project using that component, then separate into 2 sq. roots and simplify your sq. root of the perfect sq.. As in: ?seventy 2 = ?36•2 = ?36•?2 = 6?2 ?27 = ?9•3 = ?9?3 = 3?3

Answer 7

Answers:

1)sqrtx^4y^3 = sqrt ( x^(64y^3))
= x ^ (64y^3/2)
= x^ (32y^3)

2) (8sqrt5) - (sqrt20) = (8sqrt5) - sqrt(4*5)
= 8sqrt5 - 2sqrt5 = 6 sqrt (5)

3) sqrt20 1 2 sqrt 5 1
-------- - ------- = -------------- - ---------
5 sqrt5 sqrt 5 *sqrt 5 sqrt 5

2 1
= -------- - --------
sqrt 5 sqrt 5

= 1 / (sqrt 5)

Answer 8

1*
so the problem is 8(5)^(1/2) - (20)^(1/2)

to simplify these equations, you need to try to make
8(5)^(1/2) - (20)^(1/2)
....^........and..^....values the same

in this case 20 = 2^(2)*5
8(5)^(1/2) - (2^(2)*5)^(1/2)

take 2 out of the root

8(5)^(1/2) - 2(5)^(1/2)
now just subtract it and you get
6(5)^(1/2)

2*
[(20)^(1/2) / 5] - [1 / (5)^(1/2)]

the (20)^(1/2) = 2(5)^(1/2) shown on the top
and 5 = [5^(1/2)]^2
so

2(5)^(1/2) / [5^(1/2)]^2 = [(20)^(1/2) / 5]
= 2 / (5)^(1/2)

back to the equation
2 / (5)^(1/2) - 1 / (5)^(1/2)
which is
1 / (5)^(1/2)

Answer 9

1.
z=sqrt (x^4y^3)
=sqrt(x^4*y^3)
=sqrt(x^4)*sqrt(y^3)
=x^2sqrt(y^3)

2.

z = 8sqrt(5) - sqrt(20)
= 8sqrt(5) - sqrt(4*5)
= 8sqrt(5) - 2sqrt(5)
= 6sqrt(5)

3.

z = sqrt(20) - 1
---------- ----
5 sqrt(5)

z = 20^(1/2) - 1
---------- ----
5 5^(1/2)

z = 20^(1/2) - 5^(-1/2)
----------
5

z = 20^(1/2) - 5^1 * 5^(-1/2)
------------------------------
5

z = 20^(1/2) - 5^(1/2)
----------------------
5

z = sqrt(20) - sqrt(5)
----------------------
5

z = sqrt(4*5) - sqrt(5)
----------------------
5

z = 2sqrt(5) - sqrt(5)
----------------------
5

z =sqrt(5)
--------
5