SAT Math - Properties of Roots and Exponents

Right triangle $\Delta ABC$ has legs of length . What is the exact length of the hypotenuse?

$5\sqrt{2}$

$25\sqrt{2}$

$5\sqrt{5}$

$2\sqrt{5}$

Explanation

If the triangle is a right triangle, then it follows the Pythagorean Theorem. Therefore:

--->

$c= \sqrt50$

At this point, factor out the greatest perfect square from our radical:

$\sqrt{50} = \sqrt{25 \cdot 2}$

Simplify the perfect square, then repeat the process if necessary.

$\sqrt{25 \cdot 2} = 5\sqrt2$

Since is a prime number, we are finished!

What is $\sqrt{50}$ ?

$5\sqrt{2}$

$10\sqrt{2}$

$2\sqrt{5}$

Explanation

We know that 25 is a factor of 50. The square root of 25 is 5. That leaves $\sqrt{2}$ which can not be simplified further.

Which of the following is equal to $\sqrt{75}$ ?

$5\sqrt{3}$

$3\sqrt{5}$

$7.5\sqrt{10}$

Explanation

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

Which of the following is equal to $\small \sqrt{420}$ ?

$\small 2\sqrt{105}$

$\small 4\sqrt{35}$

$\small 28\sqrt{5}$

$\small 6\sqrt{35}$

$\small \small 3\sqrt{70}$

Explanation

When simplifying square roots, begin by prime factoring the number in question. For $\small 420$ , this is:

$\small 420 = 2*2*3*5*7$

Now, for each pair of numbers, you can remove that number from the square root. Thus, you can say:

$\small \small \sqrt{420}=2\sqrt{105}$

Another way to think of this is to rewrite $\small \small \sqrt{420}$ as $\small \sqrt{4}*\sqrt{105}$ . This can be simplified in the same manner.

What is the simplified (reduced) form of $\sqrt{96}$ ?

$4\sqrt{6}$

$8\sqrt{3}$

$2\sqrt{24}$

$2\sqrt{48}$

It cannot be simplified further.

Explanation

To simplify a square root, you have to factor the number and look for pairs. Whenever there is a pair of factors (for example two twos), you pull one to the outside.

Thus when you factor 96 you get
$\ \sqrt{96}=\sqrt{2\cdot 2\cdot 2\cdot2\cdot2\cdot 3}\ \=2\cdot 2\sqrt{2\cdot 3}\ \=4\sqrt{6}$

Which of the following is equivalent to $\frac{x + \sqrt{3}}{3x + \sqrt{2}}$ ?

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

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

$\frac{4x + \sqrt{5}}{3x + 2}$

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

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

Explanation

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

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

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

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

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

Simplify: $\sqrt{5000}$

$50\sqrt{2}$

$20\sqrt{5}$

$5\sqrt{20}$

$2\sqrt{50}$

$200\sqrt{5}$

Explanation

There are two ways to solve this problem. If you happen to have it memorized that is the perfect square of , then $\sqrt{5000} = \sqrt{2500 \cdot 2} = 50\sqrt 2$ gives a fast solution.

If you haven't memorized perfect squares that high, a fairly fast method can still be achieved by following the rule that any integer that ends in is divisible by , a perfect square.

$\sqrt{5000} = \sqrt{25 \cdot 200} = 5\sqrt{200}$

Now, we can use this rule again:

$5\sqrt{200} = 5\sqrt{25 \cdot 8} = 25 \sqrt {8}$

Remember that we multiply numbers that are factored out of a radical.

The last step is fairly obvious, as there is only one choice:

$25 \sqrt{8} = 25\sqrt {4 \cdot 2} = 50\sqrt 2$

Solve:

$\sqrt{20}+\sqrt{45}=?$

$5\sqrt{5}$

$\sqrt{65}$

$2\sqrt{5}$

$5\sqrt{20}$

$4\sqrt{5}$

Explanation

The trick to these problems is to simplify the radical by using the following rule: $\sqrt{ab}=\sqrt{a}*\sqrt{b}$ and $a\sqrt{b}+c\sqrt{b}=(a+c)\sqrt{b}$ Here, we need to find a common factor for the radical. This turns out to be five because $20=5*4\textup{ and }45=9*5.$ Remember, we want to include factors that are perfect squares, which are what nine and four are. Therefore, we can rewrite the equation as: $\sqrt{20}+\sqrt{45}=2\sqrt{5}+3\sqrt{5}=5\sqrt{5}.$

Each of the following is equal to $\sqrt[4]{x^3}$ for all values of EXCEPT?

$\sqrt{(\sqrt{x^3})}$

$x^{\frac{3}{4}}$

$(\sqrt[4]{x})^3$

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

Explanation

This question may look daunting, especially if you start out by trying pick values or and solving. This plan of attack will work, but you're likely going to be dealing with some messy numbers. Instead we want to recall some of our exponent and root rules.

Let's look at this answer choice:

$\sqrt{(\sqrt{x^3})}$

This double square root is the same as a fourth root. Think about it, you have a square times a square- which is the same thing as $2\times2$ . Thus, $\sqrt{(\sqrt{x^3})}=\sqrt[4]{x^3}$ so this choice can be eliminated.

Next, let's look at this answer choice:

$x^{\frac{3}{4}}$

For this choice, we need to recall our exponent rules. Remember, whenever we have a value raised to a fractional power, the denominator of that fraction is equal to the root number. In this case, . Thus, $x^{\frac{3}{4}}=\sqrt[4]{x^3}$

Now, let's look at this answer choice:

$(\sqrt[4]{x})^3$

A key rule to remember here is that order doesn't not matter when dealing with roots and powers. Thus, taking the $4^{th}}$ root of a number and then cubing it will result in the same value as cubing a number and then taking the $4^{th}}$ root .