Basic Squaring / Square Roots - SAT Math

Card 0 of 768

Question

Simplify:

√112

Answer

√112 = {√2 * √56} = {√2 * √2 * √28} = {2√28} = {2√4 * √7} = 4√7

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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

Simplify the following: (√(6) + √(3)) / √(3)

Answer

Begin by multiplying top and bottom by √(3):

(√(18) + √(9)) / 3

Note the following:

√(9) = 3

√(18) = √(9 * 2) = √(9) * √(2) = 3 * √(2)

Therefore, the numerator is: 3 * √(2) + 3. Factor out the common 3: 3 * (√(2) + 1)

Rewrite the whole fraction:

(3 * (√(2) + 1)) / 3

Simplfy by dividing cancelling the 3 common to numerator and denominator: √(2) + 1

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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:

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

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

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

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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:

√192

Answer

√192 = √2 X √96

√96 = √2 X √48

√48 = √4 X√12

√12 = √4 X √3

√192 = √(2X2X4X4) X √3

= √4X√4X√4 X √3

= 8√3

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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.

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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

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Question

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

Answer

Square both sides:

x = (32)2 = 92 = 81

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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:

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

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Question

what is

√0.0000490

Answer

easiest way to simplify: turn into scientific notation

√0.0000490= √4.9 X 10-5

finding the square root of an even exponent is easy, and 49 is a perfect square, so we can write out an improper scientific notation:

√4.9 X 10-5 = √49 X 10-6

√49 = 7; √10-6 = 10-3 this is equivalent to raising 10-6 to the 1/2 power, in which case all that needs to be done is multiply the two exponents: 7 X 10-3= 0.007

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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.

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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 :

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

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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

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Question

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

Answer

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

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

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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 .

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

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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 .

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

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