Exercises: Computing with Square Roots

With these square root exercises, you will learn to simplify root terms and compute their value.

Simplify and calculate:

$7 \sqrt{9} + 5 \sqrt{3^2+ 4^2} + 5 \sqrt{3} -3\sqrt{3}$

2
Simplify if possible:
1. $\displaystyle 5 \sqrt{3} \;+2\sqrt{3}$
  
  $5 \sqrt{3} \;+2\sqrt{3}=$
  Use the root law for adding roots.
  $(5+2)\sqrt{3}=7\sqrt{3}$
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2. $\displaystyle 6 \sqrt{4} \;-2\sqrt{4}$
  
  $6 \sqrt{4} \;-2\sqrt{4}=$
  Use the root law for subtracting roots and simplify.
  $=(6-2) \sqrt{4}\\ =4 \sqrt{4}=4 \cdot 2 = 8$
  Alternative solution
  You can also calculate the solution without the root laws:
  $6 \sqrt{4} \;-2\sqrt{4}=$
  Calculate the root.
  $6 \cdot 2 \;-2\cdot 2=$
  Calculate.
  $12 - 4 = 8$
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  Write it here 👉
3. $\displaystyle 3\sqrt{4}+3\sqrt{3}$
  
  $3\sqrt{4}+3\sqrt{3}\;$
  Here you cannot use the root law to add roots because the radicands are not equal. So you can't further simplify this term.
  What you could do is factoring out a three: $3\cdot(\sqrt{4}+\sqrt{3})\;$
  However, this is not always helpful.
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  Write it here 👉
4. $\displaystyle \sqrt{3} \cdot \sqrt{7}$
  
  $\sqrt{3} \cdot \sqrt{7}=$
  Use the root law for multiplying roots.
  $=\sqrt{3 \cdot 7} = \sqrt{21}$
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  Write it here 👉
5. $\displaystyle \sqrt{2}\cdot \sqrt{8}$
  
  $\sqrt{2}\cdot \sqrt{8}=$
  Use the root law for multiplying roots.
  $\sqrt{2 \cdot 8}=\sqrt{16}=4$
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  Write it here 👉
6. $\displaystyle \frac{\sqrt{27}}{\sqrt{3}}$
  
  $\frac{\sqrt{27}}{\sqrt{3}}=$
  Use the root law for dividing roots.
  $=\sqrt{\frac{27} {3}}=\sqrt {9}=3$
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  Write it here 👉
7. $\displaystyle \sqrt{(-27)^2}$
  
  $\sqrt{(-27)^2}$
  Use the root law for taking roots of squares.
  $= ∣ - 27 ∣ = 27$
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  Write it here 👉
8. $\displaystyle (\sqrt{2\cdot9})^2$
  
  $(\sqrt{2\cdot9})^2=(\sqrt{18})^2$
  Use the root law for taking squares of a root.
  $= 18$
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  Write it here 👉
9. $\displaystyle \sqrt7\cdot\sqrt7$
  
  $\sqrt7\cdot\sqrt7$
  Use the root law for multiplying roots.
  $= 7$
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  Write it here 👉
10. $\displaystyle 25^{\frac12}$
  
  $25^{\frac12}$
  Use the root law for converting powers into roots.
  $=\sqrt{25}=5$
  Do you have a question?
  Write it here 👉

Simplify as far as possible.

$\left(1-\sqrt3\right)\cdot\left(1+\sqrt3\right)$

$\left(\sqrt2-3\sqrt2\right)^2$

$\sqrt3\cdot\left(\frac16\sqrt{12}-3\sqrt{\frac1{27}}\right)$

$\left(2\sqrt{108}-7\sqrt{54}\right):\sqrt{27}$

$\left(\sqrt2-\sqrt{18}\right)^2$

$\left(2\sqrt7-3\right)\left(1-\sqrt{28}\right)$

$3\sqrt{63}+6\sqrt{72}-4\sqrt{28}-17\sqrt8$

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	$=$	$\displaystyle 7\sqrt{9}+5\sqrt{3^2+4^2}+5\sqrt{3}-3\sqrt{3}$
	↓	Take the root $\sqrt{9}$ .
	$=$	$\displaystyle 7\cdot3+5\sqrt{3^2+4^2}+5\sqrt{3}-3\sqrt{3}$
	↓	Compute the product.
	$=$	$\displaystyle 21+5\sqrt{3^2+4^2}+5\sqrt{3}-3\sqrt{3}$
	↓	Compute the squares.
	$=$	$\displaystyle 21+5\sqrt{9+16}+5\sqrt{3}-3\sqrt{3}$
	↓	Conclude.
	$=$	$\displaystyle 21+5\sqrt{25}+5\sqrt{3}-3\sqrt{3}$
	↓	Take the root $\sqrt{25}$ .
	$=$	$\displaystyle 21+5\cdot5+5\sqrt{3}-3\sqrt{3}$
	↓	Compute the product.
	$=$	$\displaystyle 21+25+5\sqrt{3}-3\sqrt{3}$
	$=$	$\displaystyle 46+5\sqrt{3}-3\sqrt{3}$
	↓	Conclude using the root laws.
	$=$	$\displaystyle 46+(5-3)\sqrt{3}$
	$=$

$\displaystyle \left(1-\sqrt{3}\right)\cdot\left(1+\sqrt{3}\right)$	$=$
	↓	Use the binomial formulas.
	$=$	$\displaystyle \left(1^2-\sqrt{3}^2\right)$
	↓	Square and root cancel.
	$=$	$\displaystyle \left(1-3\right)$
	$=$	$\displaystyle -2$

$\displaystyle \left(\sqrt{2}-3\sqrt{2}\right)^2$	$=$
	↓	Subtract inside the bracket
	$=$	$\displaystyle \left(-2\sqrt{2}\right)^2$
	$=$	$\displaystyle \left(-2\right)^2\cdot\left(\sqrt{2}\right)^2$
	↓	Square both parts.
	$=$	$\displaystyle 4\cdot2$
	$=$	$\displaystyle 8$

$\displaystyle \sqrt{3}\cdot\left(\frac{1}{6}\sqrt{12}-3\sqrt{\frac{1}{27}}\right)$	$=$
	↓	Pull $\frac{1}{6}$ and $3$ into the root.
	$=$	$\displaystyle \sqrt{3}\cdot\left(\sqrt{\frac{1}{36}\cdot12}-\sqrt{9\cdot\frac{1}{27}}\right)$
	↓	Multiply inside the roots.
	$=$	$\displaystyle \sqrt{3}\cdot\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{1}{3}}\right)$
	↓	Subtract inside the bracket.
	$=$	$\displaystyle \sqrt{3}\cdot0$
	$=$	$\displaystyle 0$

$\displaystyle \left(2\sqrt{108}-7\sqrt{54}\right):\sqrt{27}$	$=$
	↓	Convert the ":" into a fraction.
	$=$	$\displaystyle \frac{\left(2\sqrt{108}-7\sqrt{54}\right)}{\sqrt{27}}$
	↓	Write the fractions individually.
	$=$	$\displaystyle \frac{2\sqrt{108}}{\sqrt{27}}-\frac{7\sqrt{54}}{\sqrt{27}}$
	↓	Pull a $\sqrt{4}=2$ out of the first root.
	$=$	$\displaystyle \frac{4\sqrt{27}}{\sqrt{27}}-\frac{7\cdot\sqrt{2}\cdot\sqrt{27}}{\sqrt{27}}$
	↓	Shorten the fractions.
	$=$	$\displaystyle 4-7\sqrt{2}$

$\displaystyle \left(\sqrt{2}-\sqrt{18}\right)^2$	$=$
	↓	Apply the binomial formulas.
	$=$	$\displaystyle \sqrt{2}^2-2\cdot\sqrt{2}\cdot\sqrt{18}+\sqrt{18}^2$
	$=$	$\displaystyle 2-12+18$
	$=$	$\displaystyle 8$

$\displaystyle \left(\sqrt{2}-\sqrt{18}\right)^2$	$=$	$\displaystyle \left(\sqrt{2}-3\sqrt{2}\right)^2$
	↓	conclude
	$=$	$\displaystyle \left(-2\sqrt{2}\right)^2$
	$=$	$\displaystyle (-2)^2\cdot(\sqrt{2})^2$
	$=$	$\displaystyle 4\cdot2$
	$=$	$\displaystyle 8$

$\displaystyle \left(2\sqrt{7}-3\right)\left(1-\sqrt{28}\right)$	$=$
	↓	Pull the 2 out of the root as: $2\sqrt{7}=\sqrt{4\cdot7}$
	$=$	$\displaystyle \left(\sqrt{28}-3\right)\left(1-\sqrt{28}\right)$
	↓	Multiply out the brackets.
	$=$	$\displaystyle \sqrt{28}-28-3+3\sqrt{28}$
	↓	Simplify.
	$=$	$\displaystyle \sqrt{28}-31+3\sqrt{28}$
	$=$	$\displaystyle 4\sqrt{28}-31$

$\displaystyle 3\sqrt{63}+6\sqrt{72}-4\sqrt{28}-17\sqrt{8}$	$=$
	↓	Factorise the values under the root.
	$=$	$\displaystyle 3\sqrt{7\cdot9}+6\sqrt{2\cdot36}-4\sqrt{7\cdot4}-17\sqrt{2\cdot4}$
	↓	Take square roots.
	$=$	$\displaystyle 3\cdot3\sqrt{7}+6\cdot6\sqrt{2}-4\cdot2\sqrt{7}-34\sqrt{2}$
	$=$	$\displaystyle 9\sqrt{7}+36\sqrt{2}-8\sqrt{7}-34\sqrt{2}$
	↓	Conclude.
	$=$	$\displaystyle \sqrt{7}+2\sqrt{2}$

Exercises: Computing with Square Roots

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