How do you solve squre root of 75x^4?

Question

I need to know how do you solve a square root of 75xto the power of 4?

I need to simplify. Assuming x represents a postive number.

Answer 1

SQRT(75 x^4) =
SQRT(75) * SQRT (x^4) =
SQRT(25*3) * SQRT[(x^2)(x^2)] =
SQRT(5*5) * SQRT(3) * SQRT[(x^2)(x^2)]

By definition, the square root of the square of a number is the number. For example, the square root of (7*7) is 7.

SQRT(5*5) = 5
SQRT[(x^2)(x^2)] = x^2

Leaving us with:
5 * SQRT(3) * x^2

rearrange:
(5√3)x^2

Some would claim that the answer is "plus or minus" that number, since (-5)*(-5) = +25 same as 5*5=25 so that -5 is just as good as a square root of 25 as is +5.

+/- (5√3)x^2

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In this problem, it does not even matter if x is a negative number, as x^2 would still be positive.

Answer 2

You need to recall several facts:

1. sqrt(x * y) = sqrt(x) * sqrt(y)
2. sqrt(x^n) = x ^(n/2)

With these facts, you can factor the term 75 into a perfect square and a prime number: 75 = 3*25. Then, applying rule 1, you get sqrt(75*x^4) = sqrt(3)*sqrt(25)*sqrt(x^4).

Now, using fact two, and some common knowledge, you can reduce this to sqrt(75*x^4) = 5*sqrt(3)*x^2. This is the simplest form of the expression.

Answer 3

5x^2squqare root of 3