I stink with square roots?

Question

1)Sqrt of 56x^5y^6 over 2y^4


2)7sqrt sign3+sqrt sign 6
over
sqrt sign 6+sqrt sign 3

thanks for the help!

Answer 1

Taking a square root of some value is the same as raising that value to the power (1/2).

If you have a value that is already at some power (for example, x^4) and you raise it to some power
for example x^4 to the cube
(x^4)^3 = (x^4)(x^4)(x^4) =
(xxxx)(xxxx)(xxxx) = xxxxxxxxxxxx
= x^12
You multiply the powers (here, 4*3 = 12)

Therefore, raise the values to the power (1/2)
If you need the square root of z^8, then you take
(z^8)^(1/2) = z^(8*1/2) = z^4

When a value cannot be "rooted", then leave it as is.

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56x^5 y^5 divided by 2 y^4 =
(56/2)(x^5)(y^5/y^4)
56/2 = 28
x^5 = x^5
y^5/y^4 = yyyyy/yyyy = y (same as y^1; because 5-4=1)

SQRT(56x^5 y^5 divided by 2 y^4)
= SQRT(28 x^5 y)
=SQRT(28)*SQRT(x^5)*SQRT(y)
=SQRT(4*7)*SQRT(x*x^4)*SQRT(y)
=SQRT(4)*SQRT(7)*SQRT(x)*SQRT(x^4)*SQRT(y)

Use the following:
SQRT(4) = 2
SQRT(x^4) = (x^4)^(1/2) = x^(4/2) = x^2
SQRT(7), SQRT(x), SQRT(y) are left unchanged
(unless the question forces you to go further).

We are left with 2 x^2 * SQRT(7xy)

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The square root of a product is the product of the square roots.
SQRT(A*B) = SQRT(a)*SQRT(B)

This also works for division:
SQRT(A/B) = SQRT(A) / SQRT(B)
(as long as B is not zero)

However, this is NOT true for additions or subtractions.

SQRT(A+B) is NOT equal to SQRT(A) + SQRT(B).

In the second problem, you will have to get rid of the + signs.

I think what you have is:
[7(√3+√6)] / [(√6+√3)]

Because you are working with real numbers, you can change the order of a sum without changing its value:

A+B = B+A

Rewrite:

7 (√6 + √3)/(√6 + √3)

The (√6 + √3) cancel out, leaving you with 7.

SQRT{7 (√6 + √3)/(√6 + √3)}
=SQRT(7) * SQRT {(√6 + √3)/(√6 + √3)}
=SQRT(7)*SQRT(1)
=SQRT(7) * 1
=SQRT(7)


The square root of 7 is normally left unchanged as √7 or you can replace it with an approximation
2.6457513....