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

$5 \sqrt{3} + 2 \sqrt{3} =$
Use the root law for adding roots.
$(5 + 2) \sqrt{3} = 7 \sqrt{3}$

$6 \sqrt{4} - 2 \sqrt{4}$

$6 \sqrt{4} - 2 \sqrt{4} =$
Use the root law for subtracting roots and simplify.
$\begin{array}{l} = (6 - 2) \sqrt{4} \\ = 4 \sqrt{4} = 4 \cdot 2 = 8 \end{array}$
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.
$12 - 4 = 8$

$3 \sqrt{4} + 3 \sqrt{3}$

$3 \sqrt{4} + 3 \sqrt{3}$
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.

$\sqrt{3} \cdot \sqrt{7}$

$\sqrt{3} \cdot \sqrt{7} =$
Use the root law for multiplying roots.
$= \sqrt{3 \cdot 7} = \sqrt{21}$

$\sqrt{2} \cdot \sqrt{8}$

$\sqrt{2} \cdot \sqrt{8} =$
Use the root law for multiplying roots.
$\sqrt{2 \cdot 8} = \sqrt{16} = 4$

$\frac{\sqrt{27}}{\sqrt{3}}$

$\frac{\sqrt{27}}{\sqrt{3}} =$
Use the root law for dividing roots.
$= \sqrt{\frac{27}{3}} = \sqrt{9} = 3$

$\sqrt{(- 27)^{2}}$

$\sqrt{(- 27)^{2}}$
Use the root law for taking roots of squares.
$= | - 27 | = 27$

$(\sqrt{2 \cdot 9})^{2}$

$(\sqrt{2 \cdot 9})^{2} = (\sqrt{18})^{2}$
Use the root law for taking squares of a root.
$= 18$

$\sqrt{7} \cdot \sqrt{7}$

$\sqrt{7} \cdot \sqrt{7}$
Use the root law for multiplying roots.
$= 7$

$25^{\frac{1}{2}}$

$25^{\frac{1}{2}}$
Use the root law for converting powers into roots.
$= \sqrt{25} = 5$

Simplify as far as possible.

$(1 - \sqrt{3}) \cdot (1 + \sqrt{3})$

For this task you need the following basic knowledge: Powers
$(1 - \sqrt{3}) \cdot (1 + \sqrt{3})$ $=$
↓
Use the binomial formulas.
$=$ $(1^{2} - {\sqrt{3}}^{2})$
↓
Square and root cancel.
$=$ $(1 - 3)$
$=$ $- 2$
${(\sqrt{2} - 3 \sqrt{2})}^{2}$

For this task you need the following basic knowledge: Powers
${(\sqrt{2} - 3 \sqrt{2})}^{2}$ $=$
↓
Subtract inside the bracket
$=$ ${(- 2 \sqrt{2})}^{2}$
$=$ ${(- 2)}^{2} \cdot {(\sqrt{2})}^{2}$
↓
Square both parts.
$=$ $4 \cdot 2$
$=$ $8$
$\sqrt{3} \cdot (\frac{1}{6} \sqrt{12} - 3 \sqrt{\frac{1}{27}})$

For this task you need the following basic knowledge: Powers
$\sqrt{3} \cdot (\frac{1}{6} \sqrt{12} - 3 \sqrt{\frac{1}{27}})$ $=$
↓
Pull $\frac{1}{6}$ and $3$ into the root.
$=$ $\sqrt{3} \cdot (\sqrt{\frac{1}{36} \cdot 12} - \sqrt{9 \cdot \frac{1}{27}})$
↓
Multiply inside the roots.
$=$ $\sqrt{3} \cdot (\sqrt{\frac{1}{3}} - \sqrt{\frac{1}{3}})$
↓
Subtract inside the bracket.
$=$ $\sqrt{3} \cdot 0$
$=$ $0$

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

${(\sqrt{2} - \sqrt{18})}^{2}$

For this task you need the following basic knowledge: Powers
${(\sqrt{2} - \sqrt{18})}^{2}$ $=$
↓
Apply the binomial formulas.
$=$ ${\sqrt{2}}^{2} - 2 \cdot \sqrt{2} \cdot \sqrt{18} + {\sqrt{18}}^{2}$
$=$ $2 - 12 + 18$
$=$ $8$
Alternative Solution
${(\sqrt{2} - \sqrt{18})}^{2}$ $=$ ${(\sqrt{2} - 3 \sqrt{2})}^{2}$
↓
conclude
$=$ ${(- 2 \sqrt{2})}^{2}$
$=$ $(- 2)^{2} \cdot (\sqrt{2})^{2}$
$=$ $4 \cdot 2$
$=$ $8$
$(2 \sqrt{7} - 3) (1 - \sqrt{28})$

For this task you need the following basic knowledge: Powers
$(2 \sqrt{7} - 3) (1 - \sqrt{28})$ $=$
↓
Pull the 2 out of the root as: $2 \sqrt{7} = \sqrt{4 \cdot 7}$
$=$ $(\sqrt{28} - 3) (1 - \sqrt{28})$
↓
Multiply out the brackets.
$=$ $\sqrt{28} - 28 - 3 + 3 \sqrt{28}$
↓
Simplify.
$=$ $\sqrt{28} - 31 + 3 \sqrt{28}$
$=$ $4 \sqrt{28} - 31$
$3 \sqrt{63} + 6 \sqrt{72} - 4 \sqrt{28} - 17 \sqrt{8}$

For this task you need the following basic knowledge: Powers
$3 \sqrt{63} + 6 \sqrt{72} - 4 \sqrt{28} - 17 \sqrt{8}$ $=$
↓
Factorise the values under the root.
$=$ $3 \sqrt{7 \cdot 9} + 6 \sqrt{2 \cdot 36} - 4 \sqrt{7 \cdot 4} - 17 \sqrt{2 \cdot 4}$
↓
Take square roots.
$=$ $3 \cdot 3 \sqrt{7} + 6 \cdot 6 \sqrt{2} - 4 \cdot 2 \sqrt{7} - 34 \sqrt{2}$
$=$ $9 \sqrt{7} + 36 \sqrt{2} - 8 \sqrt{7} - 34 \sqrt{2}$
↓
Conclude.
$=$ $\sqrt{7} + 2 \sqrt{2}$

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

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

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

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

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

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

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

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

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

Exercises: Computing with Square Roots

Alternative Solution