Rationalising the denominator - Variation Theory - Free Printable

Let’s go through each problem one by one. These are all about simplifying fractions that have square roots in the denominator — a process called “rationalizing the denominator.” The goal is to get rid of the square root from the bottom (denominator) by multiplying top and bottom by the same square root.

We’ll do each pair step by step.

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First row:

1. $\frac{5}{\sqrt{2}}$
Multiply numerator and denominator by $\sqrt{2}$:
$\frac{5 \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{5\sqrt{2}}{2}$ → matches red answer.

2. $\frac{5}{\sqrt{6}}$
Multiply by $\sqrt{6}$:
$\frac{5\sqrt{6}}{6}$ → matches.

3. $\frac{5}{\sqrt{5}}$
Multiply by $\sqrt{5}$:
$\frac{5\sqrt{5}}{5} = \sqrt{5}$ → matches.

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Second row:

4. $\frac{6}{\sqrt{2}}$
Multiply by $\sqrt{2}$:
$\frac{6\sqrt{2}}{2} = 3\sqrt{2}$ → matches.

5. $\frac{5}{2\sqrt{6}}$
Multiply numerator and denominator by $\sqrt{6}$:
Numerator: $5 \cdot \sqrt{6} = 5\sqrt{6}$
Denominator: $2\sqrt{6} \cdot \sqrt{6} = 2 \cdot 6 = 12$
→ $\frac{5\sqrt{6}}{12}$ → matches.

6. $\frac{8}{\sqrt{8}}$
First, simplify $\sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2}$
So expression becomes: $\frac{8}{2\sqrt{2}} = \frac{4}{\sqrt{2}}$
Now rationalize: multiply by $\sqrt{2}$:
$\frac{4\sqrt{2}}{2} = 2\sqrt{2}$ → matches.

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Third row:

7. $\frac{6}{\sqrt{3}}$
Multiply by $\sqrt{3}$:
$\frac{6\sqrt{3}}{3} = 2\sqrt{3}$ → matches.

8. $\frac{5}{3\sqrt{6}}$
Multiply by $\sqrt{6}$:
Numerator: $5\sqrt{6}$
Denominator: $3 \cdot 6 = 18$
→ $\frac{5\sqrt{6}}{18}$ → matches.

9. $\frac{10}{\sqrt{8}}$
Simplify $\sqrt{8} = 2\sqrt{2}$ → so $\frac{10}{2\sqrt{2}} = \frac{5}{\sqrt{2}}$
Rationalize: $\frac{5\sqrt{2}}{2}$ → matches.

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Fourth row:

10. $\frac{4}{\sqrt{3}}$
Multiply by $\sqrt{3}$:
$\frac{4\sqrt{3}}{3}$ → matches.

11. $\frac{5}{10\sqrt{6}}$
Simplify first: $\frac{5}{10\sqrt{6}} = \frac{1}{2\sqrt{6}}$
Now multiply by $\sqrt{6}$:
Numerator: $1 \cdot \sqrt{6} = \sqrt{6}$
Denominator: $2 \cdot 6 = 12$
→ $\frac{\sqrt{6}}{12}$ → matches.

12. $\frac{10}{\sqrt{12}}$
Simplify $\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}$
So: $\frac{10}{2\sqrt{3}} = \frac{5}{\sqrt{3}}$
Rationalize: $\frac{5\sqrt{3}}{3}$ → matches.

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Fifth row:

13. $\frac{4}{\sqrt{6}}$
Multiply by $\sqrt{6}$:
$\frac{4\sqrt{6}}{6} = \frac{2\sqrt{6}}{3}$ → matches.

14. $\frac{5}{10\sqrt{5}}$
Simplify: $\frac{5}{10\sqrt{5}} = \frac{1}{2\sqrt{5}}$
Multiply by $\sqrt{5}$:
Numerator: $\sqrt{5}$
Denominator: $2 \cdot 5 = 10$
→ $\frac{\sqrt{5}}{10}$ → matches.

15. $\frac{10}{2\sqrt{3}}$
Simplify: $\frac{10}{2\sqrt{3}} = \frac{5}{\sqrt{3}}$
Rationalize: $\frac{5\sqrt{3}}{3}$ → matches.

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All answers shown in red are correct! Each was obtained by rationalizing the denominator — multiplying numerator and denominator by the square root in the denominator (or simplifying first if possible), then reducing the fraction if needed.

Final Answer: All given simplified forms in red are correct.