How to simplify square roots - GRE Quantitative Reasoning

easiest way to simplify: turn into scientific notation

√0.0000490= √4.9 X 10-5

finding the square root of an even exponent is easy, and 49 is a perfect square, so we can write out an improper scientific notation:

√4.9 X 10-5 = √49 X 10-6

√49 = 7; √10-6 = 10-3 this is equivalent to raising 10-6 to the 1/2 power, in which case all that needs to be done is multiply the two exponents: 7 X 10-3= 0.007

First find all of the prime factors of

So

√75 can be broken down to √25 * √3. Which simplifies to 5√3.

In order to take the square root, divide 576 by 2.

$($\frac{16}{81}$)^{1/4}$

$\frac{16^{1/4}$$$}{81^{1/4}$}

$\frac{(2cdot 2cdot 2cdot $2)^{1/4}$$}{(3cdot 3cdot 3cdot $3)^{1/4}$}

$\frac{2}{3}$

Rewrite what is under the radical in terms of perfect squares:

$x^{2}$=xcdot x

$x^{4}$$=x^{2}$cdot $x^{2}$

$x^{6}$$=x^{3}$cdot $x^{3}$

Therefore, $$\sqrt{a^{3}$$$b^{4}$$c^{5}$}= $$\sqrt{a^{2}$$$a^{1}$$b^{4}$$c^{4}$$c^{1}$$}=ab^{2}$$c^{2}$$\sqrt{ac}$.

Multiply by the conjugate and the use the formula for the difference of two squares:

$\frac{x + sqrt{3}$}{3x + $\sqrt{2}$}

$\frac{x + sqrt{3}$}{3x + $\sqrt{2}$}cdot $\frac{3x - sqrt{2}$}{3x - $\sqrt{2}$}

$$\frac{3x^{2}$$ -x $\sqrt{2}$ + 3x$\sqrt{3}$ - $$\sqrt{6}$}{(3x)^{2}$ - $($\sqrt{2}$)^{2}$}

$$\frac{3x^{2}$$ -x $\sqrt{2}$ + 3x$\sqrt{3}$ - $$\sqrt{6}$}{9x^{2}$ - 2}

We know that 25 is a factor of 50. The square root of 25 is 5. That leaves which can not be simplified further.

1. We know that , which we can separate under the square root:

2. 144 can be taken out since it is a perfect square: . This leaves us with:

This cannot be simplified any further.

You can rewrite the equation as .

This simplifies to .

Begin by multiplying top and bottom by √(3):

(√(18) + √(9)) / 3

Note the following:

√(9) = 3

√(18) = √(9 * 2) = √(9) * √(2) = 3 * √(2)

Therefore, the numerator is: 3 * √(2) + 3. Factor out the common 3: 3 * (√(2) + 1)

Rewrite the whole fraction:

(3 * (√(2) + 1)) / 3

Simplfy by dividing cancelling the 3 common to numerator and denominator: √(2) + 1

easiest way to simplify: turn into scientific notation

√0.0000490= √4.9 X 10-5

finding the square root of an even exponent is easy, and 49 is a perfect square, so we can write out an improper scientific notation:

√4.9 X 10-5 = √49 X 10-6

√49 = 7; √10-6 = 10-3 this is equivalent to raising 10-6 to the 1/2 power, in which case all that needs to be done is multiply the two exponents: 7 X 10-3= 0.007

First find all of the prime factors of

So

√75 can be broken down to √25 * √3. Which simplifies to 5√3.

In order to take the square root, divide 576 by 2.

$($\frac{16}{81}$)^{1/4}$

$\frac{16^{1/4}$$$}{81^{1/4}$}

$\frac{(2cdot 2cdot 2cdot $2)^{1/4}$$}{(3cdot 3cdot 3cdot $3)^{1/4}$}

$\frac{2}{3}$

Rewrite what is under the radical in terms of perfect squares:

$x^{2}$=xcdot x

$x^{4}$$=x^{2}$cdot $x^{2}$

$x^{6}$$=x^{3}$cdot $x^{3}$

Therefore, $$\sqrt{a^{3}$$$b^{4}$$c^{5}$}= $$\sqrt{a^{2}$$$a^{1}$$b^{4}$$c^{4}$$c^{1}$$}=ab^{2}$$c^{2}$$\sqrt{ac}$.

Multiply by the conjugate and the use the formula for the difference of two squares:

$\frac{x + sqrt{3}$}{3x + $\sqrt{2}$}

$\frac{x + sqrt{3}$}{3x + $\sqrt{2}$}cdot $\frac{3x - sqrt{2}$}{3x - $\sqrt{2}$}

$$\frac{3x^{2}$$ -x $\sqrt{2}$ + 3x$\sqrt{3}$ - $$\sqrt{6}$}{(3x)^{2}$ - $($\sqrt{2}$)^{2}$}

$$\frac{3x^{2}$$ -x $\sqrt{2}$ + 3x$\sqrt{3}$ - $$\sqrt{6}$}{9x^{2}$ - 2}

We know that 25 is a factor of 50. The square root of 25 is 5. That leaves which can not be simplified further.