help with square root problems. please help solve!?

Question

i need to express in simplest radical form.

1. the sqrt of m^2/n^3 --- (that is, m squared over n cubed together under a square root bar)

2. the sqrt of 2 PLUS the sqrt of 2/49---- (that is, the sqrt of 2 + the sqrt of 2 over 49 together under the sqrt bar)

3. (sqrt of 3 - sqrt of 5)(sqrt of 3 - sqrt of 5) --- that is, in one parenthesis the sqrt of 3 minus the sqrt of 5, next to another set of parentesis with the same thing; the sqrt of 3 minus the sqrt of 5.

Answer 1

1.)
sqrt(m^2/n^3) =
(sqrt(m^2)/sqrt(n^3)) =
sqrt(m^2n^3)/(n^3) =
(sqrt(m^2 * n(n^2))/(n^3) =
((mn)sqrt(n))/(n^3) =
(msqrt(n))/(n^2)

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2.)
sqrt(2) + sqrt(2/49) =
sqrt(2) + (1/7)sqrt(2)
(1 + (1/7))sqrt(2)
((7/7) + (1/7))sqrt(2)
(8/7)sqrt(2) or (8sqrt(2))/7

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3.)
(sqrt(3) - sqrt(5))(sqrt(3) - sqrt(5))
(3 - sqrt(15) - sqrt(15) + 5)
8 - 2sqrt(15)

Answer 2

1. The m^2 under the square root simply becomes plus or minus m (depends if your teacher cares about the + or -). The n^3 in the denominator is a bit more tricky.

Multiply √n^3 by √n^3 (as you know, whatever you do to the denominator, you must do to the numerator). You get this

±m(√n^3)/n^3

I do believe that's as simplified as it gets.

2. √(2) + √2/√49 = √(2) + √2/7 = 2√2/7

3. (√3-√5)² = √9-√(15)-√(15)+√(25) = 3 -2√(15)+5 = 8 - 2√(15)
You basically factor it out and simplify.

Answer 3

Question 1
√(m² / n³)
= m / n^(3/2)

Question 2
√2 + √(2 / 49)
= √2 + √2 / 7
=√2 / 7(7 + 1)
= 8√2 / 7

Question 3
(√3 - √5) (√3 - √5)

= 3 - 2√3 √5 + 5

= 8 - 2√15

Answer 4

i became waiting to sparkling up by employing inspection: x = 5. The sq. root of x + 4 is sqrt(9) = 3. The sq. root of x - 4 is sqrt(a million) = a million. the version is two. i presumed-approximately it as, "What 2 sq. numbers variety by employing 8 and have sq. roots that variety by employing 2?" i comprehend that 2 consecutive sq. numbers variety by employing a wierd variety equivalent to a minimum of one extra advantageous than two times the sq. root of the smaller sq.. 8 is comparable to 3 + 5, making it the sum of consecutive unusual numbers. meaning that the 1st sq. may be the sq. of one a million/2 of one below the 1st unusual variety. ((3 - a million) / 2)^2 = a million, so the 1st sq. is a million and the 2nd, 8 bigger, is 9. fixing the two x - 4 = a million or x + 4 = 9 supplies us x = 5.

Answer 5

zero