Solving Problems with Roots - SAT Math

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Question

Simplify

$\sqrt{y^{2}+13}=7$

For what value of is this equation true?

Answer

To solve for , we must first square both sides to get rid of the radical. We get $(y^{2}+13)=49$
We subtract both sides by to get the $y^{2}$ alone. $y^{2}=49-13=36$

We square root both sides to get

Answer choice and are incorrect.

Answer choice is incorrect because it was not square rooted.

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Question

Simplify

$\sqrt{5y^{2}+19}=\sqrt{259}$

For what value of is this equation true?

Answer

To solve for , we must first square both sides to get rid of the radical. We get $(5y^{2}+19)=259$ . We subtract both sides by to get the $5y^{2}$ alone. $5y^{2}=259-19=240$

We divide by to get $y^{2}$ alone $y^{2}=48$ .

We square root both sides to get $y=\pm \sqrt{48}$ Since $\pm \sqrt{48}$ is not listed as an answer choice, we simplify. The highest square root that can multiply to $\sqrt{48}$ is $\sqrt{16\times 3}$ . We take the $\sqrt{16}$ out of the radical to get $4\sqrt{3}$ .

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Question

Simplify:

$\frac{x-1}{4(\sqrt{x}+1)}$

Answer

To solve this problem we must recognize that can be broken down into $(\sqrt{x}-1)(\sqrt{x}+1)$

After breaking into $(\sqrt{x}-1)(\sqrt{x}+1)$ we see that the $(\sqrt{x}+1)$ terms in numerator and denominator cancel out.

This leaves us with $\frac{\sqrt{x}-1}{4}$

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Question

Simplify:

$\frac{5\sqrt{288}}{6}$

Answer

To solve this problem we must first simplify the radical by breaking it up into two parts $\sqrt{288}$ becomes $\sqrt{144}$ $\sqrt{2}$ then we simplify into $12\sqrt{2}$ to get $\frac{5\times 12\sqrt{2}}{6}$

We multiply $5\times 12$ to get , then divide by to get $10\sqrt{2}$

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Question

Find the value of

$7\sqrt{27}\times \sqrt{x}=21\sqrt{42}$

Answer

To solve this problem we must first simplify $\sqrt{27}$ into $\sqrt{9}\sqrt{3}$ and further into $3\sqrt{3}$

Then we can multiply $7\times 3\sqrt{3}$ to get $21\sqrt{3}$

To find we first cancel out the on both sides and then divide $\sqrt{42}$ by $\sqrt{3}$ and get $\sqrt{14}$

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Question

Find the value of

$\sqrt{5x-34}+18=29$

Answer

To solve this problem we must first subtract from both sides

$\sqrt{5x-34}=11$

Then we square both sides

Add the to both sides

Divide both sides by

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Question

Find the value of

$\frac{x-6}{4}=18-\frac{7\sqrt{x}}{4}$

Answer

To solve this problem we first multiply both sides by to get rid of the fraction

$(x-6)=4(18-\frac{7\sqrt{x}}{4})$

$x-6=72=7\sqrt{x}$

Then we add to both sides

$x=78-7\sqrt{x}$

We move $96-8\sqrt{x}$ to the left side to set the equation equal to . This way we are able to factor the equation as if it was a quadratic.

$x+7\sqrt{x}-78=0$

And now we can factor into

$(\sqrt{x}+13)(\sqrt{x}-6)=0$

Therefore the value of is

$(\sqrt{x}-6)=0, x=36$

$(\sqrt{x}+13)=0, x$ does not exist

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Question

Find the value of

$\frac{\left ( x-11 \right )}{15}=\frac{16}{3}+\frac{2\sqrt{x}}{5}$

Answer

To solve this problem we first multiply both sides by to get rid of the fraction

$(x-11)=15(\frac{16}{3}+\frac{2\sqrt{x}}5{})$

$x-11=80+6\sqrt{x}$

Then we add to both sides

$x=91+6\sqrt{x}$

We move $91-6\sqrt{x}$ to the left side to set the equation equal to . This way we are able to factor the equation as if it was a quadratic.

$x-6\sqrt{x}-91=0$

And now we can factor into

$(\sqrt{x}+7)(\sqrt{x}-13)=0$

Therefore the value of is

, does not exist

,

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Question

Find the value of

$\sqrt{18x+63}=x$

Answer

To solve this problem we must first subtract a square from both sides

$18x+63=x^{2}$

We move to the right side to set the equation equal to . This way we are able to factor the equation as if it was a quadratic.

$x^{2}-18x-63=0$

And now we can factor into

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Question

Which of the following is equivalent to $\frac{\sqrt{21}}{6}$ ?

Answer

If you try to simplify the expression given in the question, you will have a hard time…it is already simplified! However, if you look at the four answer choices you will realize that most of these contain roots in the denominator. Whenever you see a root in the denominator, you should look to rationalize that denominator. This means that you will multiply the expression by one to get rid of the root.

Consider each answer choice as you attempt to simplify each.

For choice $\frac{\sqrt{7}}2{}$ , the expression is already simplified and is not the same. At this point, your time is better spent simplifying those that need it to see if those simplified forms match.

For choice $\sqrt{\frac{7}{2}}$ , employ the "multiply by one" strategy of multiplying by the same numerator as denominator to rationalize the root. If you do so, you will multiply $\sqrt{\frac{7}{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$

$(\sqrt{\frac{7}{2}})(\frac{\sqrt{2}}{\sqrt{2}})=\frac{\sqrt{14}}{2}$ , which is no the same as $\frac{\sqrt{21}}{6}$ .

For answer choice $\sqrt{\frac{7}{12}}$ , multiply $\sqrt{\frac{7}{12}}$ by $\frac{\sqrt{12}}{\sqrt{12}}$ .

$(\sqrt{\frac{7}{12}})(\frac{\sqrt{12}}{\sqrt{12}})=\frac{\sqrt{84}}{12}$

And since $\sqrt{84}=(\sqrt{4})(\sqrt{21})$ , you can simplify the fraction:

$\frac{2\sqrt{21}}12{}=\frac{\sqrt{21}}{6}$ , which matches perfectly. Therefore, answer choice $\sqrt{\frac{7}{12}}$ is correct.

NOTE: If you want to shortcut the algebra, this problem offers you that opportunity by leveraging the answer choices along with an estimate. You can estimate that the given expression, $\frac{\sqrt{21}}6{}$ , is between $\frac{4}{6}$ and $\frac{5}{6}$ , because the $\sqrt{21}$ is between $\sqrt{16}$ (which is ) and $\sqrt{25}$ (which is ). Therefore you know you are looking for a proper fraction, a fraction in which the numerator is smaller than the denominator. Well, look at your answer choices and you will see that only answer choice $\sqrt{\frac{7}{12}}$ fits that description. So without even doing the math, you can rely on a quick estimate and know that you are correct.

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Question

If $\sqrt{10x}=10\sqrt{10}$ and $\frac{10^{-5}}{10^{-y}}=x$ , what is ?

Answer

The key to this problem is to avoid mistakes in finding with the root equation. There are a few different ways you could solve for :

1. Leverage the fact that $\sqrt{ab}=(\sqrt{a})(\sqrt{b})$ and apply that to $\sqrt{10}=\sqrt{10}\sqrt{x}$ . That means that $\sqrt{10}\sqrt{x}=10\sqrt{10}$ . Divide both sides by $\sqrt{10}$ and see that $\sqrt{x}=10$ , so .

2. Realize that $10\sqrt{10}=\sqrt{1000}$ (reverse engineering the root) and see that $\sqrt{1000}=\sqrt{10x}$ , so must equal .

However you find , you must then apply that value to the exponent expression in the second equation. Now you have $\frac{10^{-5}}{10^{-y}}=100$ . And since you're dealing with exponents, you will want to express as $10^{2}$ , meaning that you now have: $\frac{10^{-5}}{10^{-y}}=10^{2}$

Here you should deal with the negative exponents, the rule for which is that $a^{-b}=\frac{1}{a^{b}}$ . So the fraction you're given, $\frac{10^{-5}}{10^{-y}}$ , can then be transformed to $\frac{10^{y}}{10^{5}}$ .

Now you have:

$\frac{10^{y}}{10^{5}}=10^{2}$

Employing another rule of exponents, that of dividing exponents of the same base, you can transform the left-hand side to:

$\frac{10^{y}}{10^{5}}=10^{y-5}$

Since you now have everything with a base of , you can express $10^{y-5}$ as just . This then means that is the correct answer choice.

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Question

$\frac{5x^{2}}{\sqrt[3]{x^{2}}}=$

Answer

This problem emphasizes an important concept about mixing exponents and roots: when you see exponents and roots mixed, be ready to treat roots as fractional exponents. That way all of your terms are in a common form, and you can directly apply exponent rules to all terms.

Here, that means that you'll treat $\sqrt[3]{x^{2}}$ as $x^{\frac{2}{3}}$ . With that, the question then becomes:

$\frac{5x^{2}}{x^{\frac{2}{3}}}=$

Now you can use the exponent rule that $\frac{a^{b}}{a^{c}}=a^{b-c}$ (when you divide exponents of the same base, you subtract the exponents) to turn this into:

$5x^{(2-\frac{2}{3})}=5^{\frac{4}{3}}$

While you might now look to immediately convert back to the form of the answer choices (which each use radical signs to represent roots), $\sqrt[3]{x^{4}}$ is not an option. Which brings up another important point about mixing roots with exponents: generally when you're in exponent form you should stay there as long as possible, as the exponential form tends to provide you with more algebraic flexibility.

Here with $5x^{\frac{4}{3}}$ , you can pull out the $x^{1}$ (or just ) to break apart the mixed-number exponent. You can rephrase this as $5x(x^{\frac{1}{3}})$ , which then allows you to convert to an expression that looks like the answer choices. With the only fractional exponent left, $x^{\frac{1}{3}}$ , translating to $\sqrt[3]{x}$ , you can re-express the entire expression as $5x\sqrt[3]{x}$ .

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Question

$\sqrt[3]{\frac{2}{3a}}=$

Answer

This problem creates confusion for people because they are not comfortable rationalizing the cube root in the denominator. To see how to do this properly, first view the given expression like this:

$\frac{\sqrt[3]{2}}{\sqrt[3]{3a}}$

and then ask what you would need to multiply the denominator by (and thus the numerator so that you're effectively multiplying by ) to make the expression under the root sign a perfect cube in the denominator. In this case, you must multiply top and bottom by $\sqrt[3]{9a^{2}}$ as shown below:

$\frac{\sqrt[3]{2}}{\sqrt[3]{3a}}\times\frac{\sqrt[3]{9a^{2}}}{\sqrt[3]{27a^{3}}}=\frac{\sqrt[3]{18a^{2}}}{\sqrt[3]{27a^{3}}}=\frac{\sqrt[3]{18a^{2}}}{3a}$ .

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Question

$\frac{\sqrt{6}}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}}=$

Answer

To simplify this multi-denominator expression, you should first multiply by one in the form $\frac{\sqrt{6}}{\sqrt{6}}$ to eliminate the two denominators in the bottom of the expression. This multiplication results in the following:

$\frac{\sqrt{6}}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}}(\frac{\sqrt{6}}{\sqrt{6}})=\frac{6}{\frac{\sqrt{6}}{\sqrt{2}}+\frac{\sqrt{6}}{\sqrt{3}}}=\frac{6}{\sqrt{3}+\sqrt{2}}$

Now you must multiply by one again to remove the two roots in the denominator. To do that you must recognize the difference of squares and multiply the top and the bottom by the conjugate of $\sqrt{3}+\sqrt{2}$ as shown below:

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

Therefore, the correct answer is $6\sqrt{3}-6\sqrt{2}$ .

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Question

$(\sqrt{11}+\sqrt{11}+\sqrt{11})^{2}=$

Answer

All of the exponent rules deal with multiplying, rather than adding, bases. In order to turn this into a multiplication question, we count apples (or chickens, or $y=\sqrt{(11)}$ s, or whatever). How many $\sqrt{(11)}$ ’s are there here? Three. This expression can be rewritten as $(3\sqrt{(11)})^{2}$ . Now the exponent rules will apply; $(3\sqrt{(11)})^{2}=3^{2}(\sqrt{(11)})^{2}=911=99$ . The answer is .

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