ACT Math - How to simplify square roots

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!

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.

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 $\small \sqrt{420}$ ?

$\small 2\sqrt{105}$

$\small 4\sqrt{35}$

$\small 28\sqrt{5}$

$\small 6\sqrt{35}$

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

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: $\sqrt{20}+\sqrt{45}=2\sqrt{5}+3\sqrt{5}=5\sqrt{5}.$