How to add square roots - PSAT Math

Card 0 of 56

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

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Square both sides:

x = (32)2 = 92 = 81

Simplify.

$\sqrt{32}$+$\sqrt{64}$+$\sqrt{8}$

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$\sqrt{32}$+$\sqrt{64}$+$\sqrt{8}$

First step is to find perfect squares in all of our radicans.

$\sqrt{32}\rightarrow$\sqrt{16\cdot 2}$\rightarrow4$\sqrt{2}$

$$\sqrt{64}\rightarrow$\sqrt{8\cdot 8}$\rightarrow8$

$\sqrt{8}\rightarrow$\sqrt{4\cdot 2}$\rightarrow2$\sqrt{2}$

After doing so you are left with $4$\sqrt{2}$+2$\sqrt{2}$+8$

Just like fractions you can only add together coefficents with like terms under the radical.

$6$\sqrt{2}$+8$

Simplify:

$\sqrt{27}$-$\sqrt{48}$+$\sqrt{81}$

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To combine radicals, they must have the same radicand. Therefore, we must find the perfect squares in each of our square roots and pull them out.

$\sqrt{27}\rightarrow $\sqrt{93}$\rightarrow $$\sqrt{3^2$3}$\rightarrow 3$\sqrt{3}$

$\sqrt{48}\rightarrow $\sqrt{163}$\rightarrow $$\sqrt{4^2$3}$\rightarrow 4$\sqrt{3}$

Now, we plug these equivalent expressions back into our equation and simplify:

$\sqrt{27}$-$\sqrt{48}$+$\sqrt{81}$

$3\sqrt3-4\sqrt3+9$

$-\sqrt3+9$

Simplify:

$\sqrt{54}$ - $\sqrt{24}$ + $\sqrt{6}$

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Simplify each of the radicals by factoring out a perfect square:

$\sqrt{54}$ - $\sqrt{24}$ + $\sqrt{6}$

$= $\sqrt{9 \cdot 6}$ - $\sqrt{4\cdot 6}$ + $\sqrt{6}$$

$= $\sqrt{9}$ \cdot $\sqrt{6}$ - $\sqrt{4}$ \cdot $\sqrt{6}$ + $\sqrt{6}$$

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

$=\left ( 3 - 2 + 1 \right ) $\sqrt{6}$$

$= 2 $\sqrt{6}$$

Simplify the expression:

$\sqrt{45}$ + $\sqrt{125}$ + $\sqrt{175}$

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For each of the expressions, factor out a perfect square:

$\sqrt{45}$ + $\sqrt{125}$ + $\sqrt{175}$

$= $\sqrt{9 \cdot 5}$ + $\sqrt{25 \cdot 5}$ + $\sqrt{25 \cdot 7}$$

$= $\sqrt{9 }$ \cdot $\sqrt{ 5}$ + $\sqrt{25 }$ \cdot $\sqrt{ 5}$+ $\sqrt{25}$ \cdot $\sqrt{ 7}$$

$=3 $\sqrt{ 5}$ +5 $\sqrt{ 5}$+5 $\sqrt{ 7}$$

$=\left (3 +5 \right ) $\sqrt{ 5}$+5 $\sqrt{ 7}$$

$=8 $\sqrt{ 5}$+5 $\sqrt{ 7}$$

Add the square roots into one term:

$\small {$\sqrt{12}$} + {$\sqrt{75}$}$

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In order to solve this problem we need to simplfy each of the radicals. By doing this we will get two terms that have the same number under the radical which will allow us to combine the terms.

$\small {$\sqrt{12}$} + {$\sqrt{75}$}$

$\small = \small {$\sqrt{4}$} \times {$\sqrt{3}$} + {$\sqrt{25}$} \times {$\sqrt{3}$}$

$\small \small = 2 {$\sqrt{3}$} + 5 {$\sqrt{3}$}$

$\small = 7 {$\sqrt{3}$}$

Simplify:

$\sqrt{2}$+$\sqrt{2}$+$\sqrt{3}$+3$\sqrt{5}$

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Remember that you treat square roots like you do variables in the sense that you just add the like factors. In this problem, the only set of like factors is the pair of $\sqrt{2}$ values. Hence:

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

Do not try to simplify any further!

Simplify:

$\sqrt{2}$+3$\sqrt{12}$+2$\sqrt{50}$+$\sqrt{3}$

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Begin by simplifying your more complex roots:

$\sqrt{12}$= $\sqrt{223}$=2$\sqrt{3}$

$\sqrt{50}$ = $\sqrt{552}$=5$\sqrt{2}$

This lets us rewrite our expression:

$\sqrt{2}$+3$\sqrt{12}$+2$\sqrt{50}$+$\sqrt{3}$ = $\sqrt{2}$+3(2$\sqrt{3}$)+2(5$\sqrt{2}$)+$\sqrt{3}$

Do the basic multiplications of coefficients:

$\sqrt{2}$+3(2$\sqrt{3}$)+2(5$\sqrt{2}$)+$\sqrt{3}$ = $\sqrt{2}$+6$\sqrt{3}$+10$\sqrt{2}$+$\sqrt{3}$

Reorder the terms:

$\sqrt{2}$+10$\sqrt{2}$+6$\sqrt{3}$+$\sqrt{3}$

Finally, combine like terms:

$11$\sqrt{2}$+7$\sqrt{3}$$

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

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Square both sides:

x = (32)2 = 92 = 81

Simplify.

$\sqrt{32}$+$\sqrt{64}$+$\sqrt{8}$

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$\sqrt{32}$+$\sqrt{64}$+$\sqrt{8}$

First step is to find perfect squares in all of our radicans.

$\sqrt{32}\rightarrow$\sqrt{16\cdot 2}$\rightarrow4$\sqrt{2}$

$$\sqrt{64}\rightarrow$\sqrt{8\cdot 8}$\rightarrow8$

$\sqrt{8}\rightarrow$\sqrt{4\cdot 2}$\rightarrow2$\sqrt{2}$

After doing so you are left with $4$\sqrt{2}$+2$\sqrt{2}$+8$

Just like fractions you can only add together coefficents with like terms under the radical.

$6$\sqrt{2}$+8$

Simplify:

$\sqrt{27}$-$\sqrt{48}$+$\sqrt{81}$

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To combine radicals, they must have the same radicand. Therefore, we must find the perfect squares in each of our square roots and pull them out.

$\sqrt{27}\rightarrow $\sqrt{93}$\rightarrow $$\sqrt{3^2$3}$\rightarrow 3$\sqrt{3}$

$\sqrt{48}\rightarrow $\sqrt{163}$\rightarrow $$\sqrt{4^2$3}$\rightarrow 4$\sqrt{3}$

Now, we plug these equivalent expressions back into our equation and simplify:

$\sqrt{27}$-$\sqrt{48}$+$\sqrt{81}$

$3\sqrt3-4\sqrt3+9$

$-\sqrt3+9$

Simplify:

$\sqrt{54}$ - $\sqrt{24}$ + $\sqrt{6}$

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Simplify each of the radicals by factoring out a perfect square:

$\sqrt{54}$ - $\sqrt{24}$ + $\sqrt{6}$

$= $\sqrt{9 \cdot 6}$ - $\sqrt{4\cdot 6}$ + $\sqrt{6}$$

$= $\sqrt{9}$ \cdot $\sqrt{6}$ - $\sqrt{4}$ \cdot $\sqrt{6}$ + $\sqrt{6}$$

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

$=\left ( 3 - 2 + 1 \right ) $\sqrt{6}$$

$= 2 $\sqrt{6}$$

Simplify the expression:

$\sqrt{45}$ + $\sqrt{125}$ + $\sqrt{175}$

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For each of the expressions, factor out a perfect square:

$\sqrt{45}$ + $\sqrt{125}$ + $\sqrt{175}$

$= $\sqrt{9 \cdot 5}$ + $\sqrt{25 \cdot 5}$ + $\sqrt{25 \cdot 7}$$

$= $\sqrt{9 }$ \cdot $\sqrt{ 5}$ + $\sqrt{25 }$ \cdot $\sqrt{ 5}$+ $\sqrt{25}$ \cdot $\sqrt{ 7}$$

$=3 $\sqrt{ 5}$ +5 $\sqrt{ 5}$+5 $\sqrt{ 7}$$

$=\left (3 +5 \right ) $\sqrt{ 5}$+5 $\sqrt{ 7}$$

$=8 $\sqrt{ 5}$+5 $\sqrt{ 7}$$

Add the square roots into one term:

$\small {$\sqrt{12}$} + {$\sqrt{75}$}$

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In order to solve this problem we need to simplfy each of the radicals. By doing this we will get two terms that have the same number under the radical which will allow us to combine the terms.

$\small {$\sqrt{12}$} + {$\sqrt{75}$}$

$\small = \small {$\sqrt{4}$} \times {$\sqrt{3}$} + {$\sqrt{25}$} \times {$\sqrt{3}$}$

$\small \small = 2 {$\sqrt{3}$} + 5 {$\sqrt{3}$}$

$\small = 7 {$\sqrt{3}$}$

Simplify:

$\sqrt{2}$+$\sqrt{2}$+$\sqrt{3}$+3$\sqrt{5}$

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Remember that you treat square roots like you do variables in the sense that you just add the like factors. In this problem, the only set of like factors is the pair of $\sqrt{2}$ values. Hence:

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

Do not try to simplify any further!

Simplify:

$\sqrt{2}$+3$\sqrt{12}$+2$\sqrt{50}$+$\sqrt{3}$

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Begin by simplifying your more complex roots:

$\sqrt{12}$= $\sqrt{223}$=2$\sqrt{3}$

$\sqrt{50}$ = $\sqrt{552}$=5$\sqrt{2}$

This lets us rewrite our expression:

$\sqrt{2}$+3$\sqrt{12}$+2$\sqrt{50}$+$\sqrt{3}$ = $\sqrt{2}$+3(2$\sqrt{3}$)+2(5$\sqrt{2}$)+$\sqrt{3}$

Do the basic multiplications of coefficients:

$\sqrt{2}$+3(2$\sqrt{3}$)+2(5$\sqrt{2}$)+$\sqrt{3}$ = $\sqrt{2}$+6$\sqrt{3}$+10$\sqrt{2}$+$\sqrt{3}$

Reorder the terms:

$\sqrt{2}$+10$\sqrt{2}$+6$\sqrt{3}$+$\sqrt{3}$

Finally, combine like terms:

$11$\sqrt{2}$+7$\sqrt{3}$$

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

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Square both sides:

x = (32)2 = 92 = 81

Simplify.

$\sqrt{32}$+$\sqrt{64}$+$\sqrt{8}$

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$\sqrt{32}$+$\sqrt{64}$+$\sqrt{8}$

First step is to find perfect squares in all of our radicans.

$\sqrt{32}\rightarrow$\sqrt{16\cdot 2}$\rightarrow4$\sqrt{2}$

$$\sqrt{64}\rightarrow$\sqrt{8\cdot 8}$\rightarrow8$

$\sqrt{8}\rightarrow$\sqrt{4\cdot 2}$\rightarrow2$\sqrt{2}$

After doing so you are left with $4$\sqrt{2}$+2$\sqrt{2}$+8$

Just like fractions you can only add together coefficents with like terms under the radical.

$6$\sqrt{2}$+8$

Simplify:

$\sqrt{27}$-$\sqrt{48}$+$\sqrt{81}$

Tap to see back →

To combine radicals, they must have the same radicand. Therefore, we must find the perfect squares in each of our square roots and pull them out.

$\sqrt{27}\rightarrow $\sqrt{93}$\rightarrow $$\sqrt{3^2$3}$\rightarrow 3$\sqrt{3}$

$\sqrt{48}\rightarrow $\sqrt{163}$\rightarrow $$\sqrt{4^2$3}$\rightarrow 4$\sqrt{3}$

Now, we plug these equivalent expressions back into our equation and simplify:

$\sqrt{27}$-$\sqrt{48}$+$\sqrt{81}$

$3\sqrt3-4\sqrt3+9$

$-\sqrt3+9$

Simplify:

$\sqrt{54}$ - $\sqrt{24}$ + $\sqrt{6}$

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Simplify each of the radicals by factoring out a perfect square:

$\sqrt{54}$ - $\sqrt{24}$ + $\sqrt{6}$

$= $\sqrt{9 \cdot 6}$ - $\sqrt{4\cdot 6}$ + $\sqrt{6}$$

$= $\sqrt{9}$ \cdot $\sqrt{6}$ - $\sqrt{4}$ \cdot $\sqrt{6}$ + $\sqrt{6}$$

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

$=\left ( 3 - 2 + 1 \right ) $\sqrt{6}$$

$= 2 $\sqrt{6}$$