# Solutions to Simplify Radicals Questions

Solutions with detailed explanations to questions on simplifying radicals are presented.

   $\require{cancel}$ $\newcommand\ccancel[2][black]{\color{#1}{\xcancel{\color{black}{#2}}}}$

In the solutions below, we use the product rule of radicals given by
$\sqrt{x \times y} = \sqrt{x } \sqrt{y}$

1. Simplify the expression $2 \sqrt{50} + 12 \sqrt{8}$.
Solution
Use the fact that $50 = 2 \times 25$ and $8 = 2 \times 4$ to rewrite the given expressions as follows
$\quad 2 \sqrt{50} + 12 \sqrt{8} = 2 \sqrt {2 \times 25} + 12 \sqrt{2 \times 4}$
Use the product rule of radicals $\sqrt{x \times y} = \sqrt{x } \sqrt{y}$ to rewrite the above as
$\quad = 2 \sqrt 2 \sqrt{25} + 12 \sqrt 2 \sqrt 4$
Use the fact that $\sqrt{25} = 5$ and $\sqrt 4 = 2$ to rewrite the above as
$\quad = 2 \times 5 \times \sqrt 2 + 12 \times 2 \times \sqrt 2$
Simplify
$\quad = 10 \sqrt 2 + 24 \sqrt 2$
Factor $\sqrt 2$ out and simplify
$\quad = (10 + 24) \sqrt 2 = 34 \sqrt 2$

2. Simplify the expression $\sqrt{27} - \sqrt{300}$.
Solution
Knowing that $27 = 3 \times 9$ and $300 = 3 \times 100$, we write
$\sqrt{27} - \sqrt{300} = \sqrt{3 \times 9} - \sqrt {3 \times 100}$
Use the rule for radicals of radicals $\sqrt{x \times y} = \sqrt{x } \sqrt{y}$ to rewrite the above as
$= \sqrt{3} \sqrt{9} - \sqrt {3} \sqrt{100}$
Knowing that $\sqrt{9} = 3$ and $\sqrt{100} = 10$, the above may be simplified to
$= 3 \sqrt 3 - 10 \sqrt3 = (3 - 10 )\sqrt 3 = - 7 \sqrt 3$

3. Simplify the expression $- 2 \sqrt{16y} + 10 \sqrt y$.
Solution
Use the product rule of radicals $\sqrt{x \times y} = \sqrt{x } \sqrt{y}$ to rewrite $\sqrt{16y}$ as $\sqrt {16} \sqrt y$ and substitute in the given expression
$- 2 \sqrt{16y} + 10 \sqrt y = - 2 \sqrt {16} \sqrt y + 10 \sqrt y$
Use the fact that $\sqrt {16} = 4$ and substitute in the above
$= - 2 \times 4 \sqrt y + 10 \sqrt y$
$= - 8 \sqrt y + 10 \sqrt y$
Factor $\sqrt y$ out and simplify
$= (-8 + 10) \sqrt y = 2 \sqrt y$

4. Simplify the expression $2 \sqrt{x + 1} + 3 \sqrt{16x + 16}$.
Solution
Factor 16 out in $16x + 16$
$2 \sqrt(x + 1) + 3 \sqrt(16x + 16) = 2 \sqrt{x + 1} + 3 \sqrt{16(x + 1) }$
Use the product rule of radicals to write $\sqrt{16(x + 1)}$ as $\sqrt{16} \sqrt {x-1} = 4 \sqrt {x-1}$ and substitute in the above
$= 2 \sqrt{x + 1} + 3 \times 4 \times \sqrt{x + 1}$
Simplify
$= 2 \sqrt{x + 1} + 12 \sqrt{x + 1}$
Factor $\sqrt{x + 1}$ and simplify
$= ( 2 + 12 ) \sqrt{x + 1} = 14 \sqrt{x + 1}$

5. $2 \sqrt 3 + 4 \sqrt{12} + 3 \sqrt{48} =$
Solution
Write the radicands as product of numbers whose square roots are integers: $12 = 4 \times 3$ and $48 = 3 \times 16$
$2 \sqrt3 + 4 \sqrt{12} + 3 \sqrt{48} = 2 \sqrt3 + 4 \sqrt{3 \times 4} + 3 \sqrt{3 \times 16}$
Use the product rule of radicals to write the above as
$= 2 \sqrt3 + 4 \sqrt3 \sqrt4 + 3 \sqrt3 \sqrt{16}$
Simplify $\sqrt4$ and $\sqrt{16}$
$= 2 \sqrt3 + 8 \sqrt3 + 12 \sqrt3$
Factor $\sqrt3$ and rewrite the above as
$= (2 + 8 + 12) \sqrt3$
Simplify
$= 22 \sqrt3$

6. Rewrite the expression $\dfrac {\sqrt3 + \sqrt{12}} {\sqrt 3 - \sqrt{12}}$ without radicals.
Solution
We rewrite the given expression using the fact that $\sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \sqrt{3} = 2 \sqrt 3$
$\dfrac {\sqrt3 + \sqrt{12}} {\sqrt 3 - \sqrt{12}} = \dfrac {\sqrt3 + 2 \sqrt{3}} {\sqrt 3 - 2 \sqrt{3}}$
Group terms in the numerator and denominator
$= \dfrac {(1+2) \sqrt{3}} {(1-2) \sqrt{3}} = \dfrac{3 \sqrt 3}{-\sqrt 3}$
Divide numerator and denominator by $\sqrt 3$ and simplify (cancel)
$= - 3$

7. Simplify the expression $5 \sqrt x + 6 \sqrt {9x} - 10 \sqrt {16x}$.
Solution
Use the product rule of radicals to rewrite the above as
$5 \sqrt x + 6 \sqrt {9x} - 10 \sqrt {16x} = 5 \sqrt x + 6 \sqrt9 \sqrt x - 10 \sqrt{16} \sqrt x$
Simplify $\sqrt 9$ and $\sqrt{16}$
$= 5 \sqrt x + 6 \times 3 \sqrt x - 10 \times 4 \sqrt x$
Simplify
$= 5 \sqrt x + 18 \sqrt x - 40 \sqrt x$
Factor $\sqrt x$ and simplify
$= (5 + 18 - 40) \sqrt x = -17 \sqrt x$

8. $2 \sqrt {27} + 2 \sqrt{75} =$
Solution
Write 27 and 75 as products of numbers whose square roots are integers: $27 = 3 \times 9$ and $75 = 3 \times 25$
$2 \sqrt {27} + 2 \sqrt{75} = 2 \sqrt {3 \times 9} + 2 \sqrt{3 \times 25}$
Use the product rule of radicals
$= 2 \sqrt 3 \sqrt 9 + 2 \sqrt{3} \sqrt{ 25}$
Simplify using $\sqrt 9 = 3$ and $\sqrt{ 25} = 5$
$= 6 \sqrt 3 + 10 \sqrt{3}$
Factor $\sqrt{3}$ out and simplify
$= (6 + 10) \sqrt{3} = 16 \sqrt{3}$

9. $\sqrt {10^3} + \sqrt {10^5} =$
Solution
Write $10^3$ and $10^5$ as products of numbers whose square roots are integers: $10^3 = 10 \times 100$ and $10^5 = 10 \times 10000$
$\sqrt {10^3} + \sqrt {10^5} = \sqrt {10 \times 100} + \sqrt{10 \times 10000}$
Use the product rule of radicals
$= \sqrt {10} \sqrt {100} + \sqrt{10} \sqrt{ 10000}$
Simplify using $\sqrt 100 = 10$ and $\sqrt{ 10000} = 100$
$= 10 \sqrt 10 + 100 \sqrt{10}$
Factor $\sqrt{10}$ out and simplify
$= (10 + 100) \sqrt{10} = 110 \sqrt{10}$

10. Simplify and rewrite the expression $\sqrt 8 \sqrt 3 \sqrt 6$ without radicals.
Solution
Write $8$ as $2 \times 4$
$\sqrt 8 \sqrt 3 \sqrt 6 = \sqrt{ 4 \times 2 } \sqrt 3 \sqrt 6$
Use the product rule of radicals on $\sqrt{ 4 \times 2 }$ and simplify
$= \sqrt 4 \sqrt 2 \sqrt 3 \sqrt 6 = 2 \sqrt 2 \sqrt 3 \sqrt 6$
Use product rule of radicals to write $\sqrt 2 \sqrt 3$ as $\sqrt 6$ and substitute
$= 2 \sqrt 6 \sqrt 6$
Use product rule of radicals to write $\sqrt 6 \sqrt 6 = \sqrt {36} = 6$ and simplify
$= 2 \times 6 = 12$