Basic Squaring / Square Roots - GRE Quantitative Reasoning

Begin by breaking down each of the roots in question. Often, for the GRE, your answer arises out of doing such basic "simplification moves".

Quantity A

This is the same as:

, which can be reduced to:

Quantity B

This is the same as:

, which can be reduced to:

Thus, at the end of working through the proper math, you realize that the two values are equal!

Begin by factoring out each of the radicals:

For the first two radicals, you can factor out a or :

The other root values cannot be simply broken down. Now, combine the factors with :

This is your simplest form.

Begin by getting your terms onto the left side of the equation and your numeric values onto the right side of the equation:

Next, you can combine your radicals. You do this merely by subtracting their respective coefficients:

Now, square both sides:

Solve by dividing both sides by :

To solve this problem, we must realize that the only way to add or subtract square roots is if the number under to square root is equivalent to each other. Therefore we must find a way to simplify each square root.

First we attempt to simplify the first term,

We break apart the number under the square root and find

Simplifying

Therefore we know that in order to try and simplify the other terms, the number under the square root has to be 3. By removing from the other terms in the equation, we will attempt to see if they can be simplified as well.

For the second term,

Finally for the last term,

Our new equation becomes

Once all number under the square roots become the same, we can treat it as simple addition/subtraction and solve.

Take each fraction separately first:

(2√3)/(√2) = \[(2√3)/(√2)\] * \[(√2)/(√2)\] = (2 * √3 * √2)/(√2 * √2) = (2 * √6)/2 = √6

Similarly:

(4√2)/(√3) = \[(4√2)/(√3)\] * \[(√3)/(√3)\] = (4√6)/3 = (4/3)√6

Now, add them together:

√6 + (4/3)√6 = (3/3)√6 + (4/3)√6 = (7/3)√6

Begin by breaking down each of the roots in question. Often, for the GRE, your answer arises out of doing such basic "simplification moves".

Quantity A

This is the same as:

, which can be reduced to:

Quantity B

This is the same as:

, which can be reduced to:

Thus, at the end of working through the proper math, you realize that the two values are equal!

Begin by factoring out each of the radicals:

For the first two radicals, you can factor out a or :

The other root values cannot be simply broken down. Now, combine the factors with :

This is your simplest form.

Begin by getting your terms onto the left side of the equation and your numeric values onto the right side of the equation:

Next, you can combine your radicals. You do this merely by subtracting their respective coefficients:

Now, square both sides:

Solve by dividing both sides by :

To solve this problem, we must realize that the only way to add or subtract square roots is if the number under to square root is equivalent to each other. Therefore we must find a way to simplify each square root.

First we attempt to simplify the first term,

We break apart the number under the square root and find

Simplifying

Therefore we know that in order to try and simplify the other terms, the number under the square root has to be 3. By removing from the other terms in the equation, we will attempt to see if they can be simplified as well.

For the second term,

Finally for the last term,

Our new equation becomes

Once all number under the square roots become the same, we can treat it as simple addition/subtraction and solve.

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.

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.