Basic Squaring / Square Roots - SAT Math

Card 0 of 20

Question

Simplify in radical form:

$\sqrt{507}+\sqrt{12}$

Answer

To simplify, break down each square root into its component factors:

$\sqrt{507}+\sqrt{12}$

$\sqrt{169\cdot 3}+\sqrt{4\cdot 3}$

$\sqrt{13\cdot13\cdot3}:+\sqrt{2\cdot 2\cdot 3}$

You can remove pairs of factors and bring them outside the square root sign. At this point, since each term shares $\sqrt3$ , you can add them together to yield the final answer:

$13\sqrt{3}+2\sqrt{3}=15\sqrt{3}$

Compare your answer with the correct one above

Question

Simplify: $\frac{2\sqrt{3}}{\sqrt{2}}+\frac{4\sqrt{2}}{\sqrt{3}}$

Answer

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

Compare your answer with the correct one above

Question

If $\sqrt{x}=3^2$ what is $x$ ?

Answer

Square both sides:

x = (32)2 = 92 = 81

Compare your answer with the correct one above

Question

Simplify the following expression: $\sqrt{60}+\sqrt{40}+\sqrt{10}$

Answer

Begin by factoring out each of the radicals:

$\sqrt{60}+\sqrt{40}+\sqrt{10}=\sqrt{415}+\sqrt{410}+\sqrt{10}$

For the first two radicals, you can factor out a $\sqrt{4}$ or :

$2\sqrt{15}+2\sqrt{10}+\sqrt{10}$

The other root values cannot be simply broken down. Now, combine the factors with $\sqrt{10}$ :

$2\sqrt{15}+2\sqrt{10}+\sqrt{10}=2\sqrt{15}+3\sqrt{10}$

This is your simplest form.

Compare your answer with the correct one above

Question

Solve for .

Note, $x\ge0$ :

$15\sqrt{x}-10=4\sqrt{x}+4$

Answer

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

$15\sqrt{x}-4\sqrt{x}=4+10$

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

$11\sqrt{x}=14$

Now, square both sides:

$(11\sqrt{x})^2=(14)^2$

$(11)^2(\sqrt{x})^2=(14)^2$

Solve by dividing both sides by :

$x=\frac{196}{121}$

Compare your answer with the correct one above

Question

(√27 + √12) / √3 is equal to

Answer

√27 is the same as 3√3, while √12 is the same as 2√3.

3√3 + 2√3 = 5√3

(5√3)/(√3) = 5

Compare your answer with the correct one above

Question

Simplify: $\frac{\sqrt{210}}{\sqrt{6}}$

Answer

When dividing square roots, we divide the numbers inside the radical. Simplify if necessary.

$\frac{\sqrt{210}}{\sqrt{6}}=\frac{\sqrt{35}*\sqrt{6}}{\sqrt{6}}=\sqrt{35}$

Compare your answer with the correct one above

Question

Simplify: $\frac{\sqrt{432}}{\sqrt{36}}$

Answer

When dividing square roots, we divide the numbers inside the radical. Simplify if necessary.

$\frac{\sqrt{432}}{\sqrt{36}}=\frac{\sqrt{36}\sqrt{12}}{\sqrt{36}}=\sqrt{12}$

Let's simplify this even further by factoring out a .

$\sqrt{12}=\sqrt{4}\sqrt{3}=2\sqrt{3}$

Compare your answer with the correct one above

Question

Simplify: $\frac{\sqrt{1200}}{\sqrt{8}}$

Answer

When dividing square roots, we divide the numbers inside the radical. Simplify if necessary.

$\frac{\sqrt{1200}}{\sqrt{8}}=\frac{\sqrt{8}\sqrt{150}}{\sqrt{8}}=\sqrt{150}$
Let's simplify this even further by factoring out a .

$\sqrt{150}=\sqrt{25}\sqrt{6}=5\sqrt{6}$

Compare your answer with the correct one above

Question

Solve for $\dpi{100} x$ :

$x\sqrt{45}+x\sqrt{72}=\sqrt{18}$

Answer

$x\sqrt{45}+x\sqrt{72}=\sqrt{18}$

Notice how all of the quantities in square roots are divisible by 9

$x\sqrt{9\times 5}+x\sqrt{9\times 8}=\sqrt{9\times 2}$

$x\sqrt{9}\sqrt{5}+x\sqrt{9}\sqrt{4\times 2}=\sqrt{9}\sqrt{2}$

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

$3x\sqrt{5}+6x\sqrt{2}=3\sqrt{2}$

$x(3\sqrt{5}+6\sqrt{2})=3\sqrt{2}$

$x=\frac{3\sqrt{2}}{3\sqrt{5}+6\sqrt{2}}$

Simplifying, this becomes

$x=\frac{\sqrt{2}}{\sqrt{5}+2\sqrt{2}}$

Compare your answer with the correct one above

Question

Solve for :

$x\sqrt{27} + \sqrt{63}=9$

Answer

In order to solve for , first note that all of the square root terms on the left side of the equation have a common factor of 9 and 9 is a perfect square:

$x\sqrt{27} + \sqrt{63}=9$

$x\sqrt{9\times 3} + \sqrt{9\times 7}=9$

$3x\sqrt{3} + 3\sqrt{7}=9$

Simplifying, this becomes:

$3(x\sqrt{3} + \sqrt{7})=9$

$x\sqrt{3} + \sqrt{7}=3$

$x\sqrt{3} =3 -\sqrt{7}$

$x =\frac{3 -\sqrt{7}}{\sqrt{3}}$

Compare your answer with the correct one above

Question

Solve for :

$x\sqrt{72}+x\sqrt{108}=12$

Answer

$x\sqrt{72}+x\sqrt{108}=12$

Note that all of the square root terms share a common factor of 36, which itself is a square of 6:

$x\sqrt{36\times 2}+x\sqrt{36\times 3}=12$

$6x\sqrt{2}+6x\sqrt{3}=12$

Factoring from both terms on the left side of the equation:

$6x(\sqrt{2}+\sqrt{3})=12$

$6x=\frac{12}{\sqrt{2}+\sqrt{3}}$

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

Compare your answer with the correct one above

Question

Solve for :

$2x\sqrt{12}+3x\sqrt{28}=18$

Answer

$2x\sqrt{12}+3x\sqrt{28}=18$

Note that both $\sqrt{12}$ and $\sqrt{28}$ have a common factor of and is a perfect square:

$2x\sqrt{4\times 3}+3x\sqrt{4\times 7}=18$

$2x\cdot2 \sqrt3 +3x\cdot2\sqrt7=18$

$4x\sqrt{3}+6x\sqrt{7}=18$

From here, we can factor out of both terms on the lefthand side

$2x(2\sqrt{3}+3\sqrt{7})=18$

$2x(2\sqrt{3}+3\sqrt{7})=18$

$x=\frac{18}{2(2\sqrt{3}+3\sqrt{7})}$

$x=\frac{9}{2\sqrt{3}+3\sqrt{7}}$

Compare your answer with the correct one above

Question

Which of the following is equivalent to:

$\sqrt{210}+\sqrt{55}$ ?

Answer

To begin with, factor out the contents of the radicals. This will make answering much easier:

$\sqrt{210}+\sqrt{55}=\sqrt{2357}+\sqrt{511}$

They both have a common factor . This means that you could rewrite your equation like this:

$\sqrt{5}\sqrt{237}+\sqrt{5}\sqrt{11}$

This is the same as:

$\sqrt{5}\sqrt{42}+\sqrt{5}\sqrt{11}$

These have a common $\sqrt{5}$ . Therefore, factor that out:

$\sqrt{5}\sqrt{42}+\sqrt{5}\sqrt{11}=\sqrt{5}(\sqrt{42}+\sqrt{11})$

Compare your answer with the correct one above

Question

Simplify:

$\sqrt{15}-\sqrt{20}+\sqrt{35}$

Answer

These three roots all have a in common; therefore, you can rewrite them:

$\sqrt{15}-\sqrt{20}+\sqrt{35}=\sqrt{35}-\sqrt{45}+\sqrt{75}$

Now, this could be rewritten:

$\sqrt{35}-\sqrt{45}+\sqrt{75}=\sqrt{5}\sqrt{3}-\sqrt{5}\sqrt{4}+\sqrt{5}\sqrt{7}$

Now, note that $\sqrt{4}=2$

Therefore, you can simplify again:

$\sqrt{5}\sqrt{3}-\sqrt{5}\sqrt{4}+\sqrt{5}\sqrt{7}=\sqrt{5}\sqrt{3}-2\sqrt{5}+\sqrt{5}\sqrt{7}$

Now, that looks messy! Still, if you look carefully, you see that all of your factors have $\sqrt{5}$ ; therefore, factor that out:

$\sqrt{5}\sqrt{3}-2\sqrt{5}+\sqrt{5}\sqrt{7}=\sqrt{5}(\sqrt{3}-2+\sqrt{7})$

This is the same as:

$\sqrt{5}(\sqrt{3}+\sqrt{7}-2)$

Compare your answer with the correct one above

Question

Solve for :

$2x\sqrt{128}-3x\sqrt{192}=16$

Answer

Examining the terms underneath the radicals, we find that and have a common factor of . itself is a perfect square, being the product of and . Hence, we recognize that the radicals can be re-written in the following manner:

$\sqrt{128}=\sqrt{64} \sqrt{2}$ , and $\sqrt{192}=\sqrt{64} \sqrt{3}$ .

The equation can then be expressed in terms of these factored radicals as shown:

$2x\sqrt{128}-3x\sqrt{192}=16$

$\Rightarrow 2x(\sqrt64 \sqrt2)-3x(\sqrt{64} \sqrt3)=16$

$\Rightarrow 2x(8)\sqrt2-3x(8)\sqrt3=16$

$\Rightarrow 16x\sqrt2-24x\sqrt3=16$

Factoring the common term from the lefthand side of this equation yields

$8x (2\sqrt2-3x\sqrt3)=16$

Divide both sides by the expression in the parentheses:

$8x=\frac{16}{2\sqrt2-3x\sqrt3}$

Divide both sides by to yield by itself on the lefthand side:

$x=\frac{16}{8(2\sqrt2-3x\sqrt3)}$

Simplify the fraction on the righthand side by dividing the numerator and denominator by :

$x=\frac{2}{2\sqrt{2}-3\sqrt{3}}$

This is the solution for the unknown variable that we have been required to find.

Compare your answer with the correct one above

Question

If m and n are postive integers and 4m = 2n, what is the value of m/n?

Answer

22 = 4. Also, following the rules of exponents, 41 = 1.

One can therefore say that m = 1 and n = 2.

The question asks to solve for m/n. Since m = 1 and n = 2, m/n = 1/2.

Compare your answer with the correct one above

Question

Simplify:

$\sqrt[2]{24,300}$

Answer

$\sqrt[2]{24,300}$

When simplifying the square root of a number that may not have a whole number root, it's helpful to approach the problem by finding common factors of the number inside the radicand. In this case, the number is 24,300.

What are the factors of 24,300?

24,300 can be factored into:

$24,300: 243\cdot 100$

When there are factors that appear twice, they may be pulled out of the radicand. For instance, 100 is a multiple of 24,300. When 100 is further factored, it is $10^{2}$ (or 10x10). However, 100 wouldn't be pulled out of the radicand, but the square root of 100 because the square root of 24,300 is being taken. The 100 is part of the24,300. This means that the problem would be rewritten as: $10\sqrt[2]{243}$

But 243 can also be factored: $24,300: (3\cdot 81)\cdot100$
$24,300: {\color{Orange} 9}\cdot{\color{Orange} 9}\cdot {\color{Cyan} 10}\cdot {\color{cyan} 10}\cdot 3$
Following the same principle as for the 100, the problem would become
$9\cdot 10\sqrt[2]{3}$ because there is only one factor of 3 left in the radicand. If there were another, the radicand would be lost and it would be 9103.
9 and 10 may be multiplied together, yielding the final simplified answer of
$90\sqrt[2]{3}$

Compare your answer with the correct one above

Question

$\sqrt{180} + \sqrt{125} = ?$

Answer

To solve the equation $\sqrt{180} + \sqrt{125} = ?$ , we can first factor the numbers under the square roots.

$\sqrt{2\cdot 2\cdot 3\cdot 3\cdot 5} + \sqrt{5\cdot 5\cdot 5} = ?$

When a factor appears twice, we can take it out of the square root.

$2\cdot 3\sqrt{5} + 5\sqrt{5} = ?$

$6\sqrt{5} + 5\sqrt{5} = ?$

Now the numbers can be added directly because the expressions under the square roots match.

$6\sqrt{5} + 5\sqrt{5} = 11\sqrt{5}$

Compare your answer with the correct one above

Question

How many integers from 20 to 80, inclusive, are NOT the square of another integer?

Answer

First list all the integers between 20 and 80 that are squares of another integer:

52 = 25

62 = 36

72 = 49

82 = 64

In total, there are 61 integers from 20 to 80, inclusive. 61 – 4 = 57

Compare your answer with the correct one above

Tap the card to reveal the answer