Factoring Common Factors of Squares and Square Roots - SAT Math

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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

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

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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:

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

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

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

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

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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

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

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

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

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

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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:

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

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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:

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

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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

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

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

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