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Rationalising The Denominator

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Rationalising The Denominator – Introduction

To rationalise the denominator in GCSE Maths, you can follow these steps:
 Identify the denominator of the fraction that needs to be rationalised.
 Multiply both the numerator and denominator by the conjugate of the denominator.
 Simplify the resulting expression by applying the distributive property and combining like terms.
 The denominator should now be rationalised, without any square roots or other irrational values. 

Example:
 Consider a fraction with a denominator of √3. To rationalise it, we multiply both the numerator and denominator by the conjugate of √3, which is also √3. This results in (√3)(√3) in the denominator, which simplifies to 3. Therefore, the denominator is rationalised.

Rationalising The Denominator - Examples 1 to 4

\begin{aligned} \frac{10}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} & =\frac{10 \sqrt{5}}{5} \\ & =2 \sqrt{5} \end{aligned}

\begin{aligned} \frac{18}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} & =\frac{18 \sqrt{6}}{6} \\ & =3 \sqrt{6} \end{aligned}

\begin{aligned} \frac{70}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} & =\frac{70 \sqrt{2}}{2} \\ & =35 \sqrt{2} \end{aligned}

\begin{aligned} \frac{20}{\sqrt{10}} \times \frac{\sqrt{10}}{\sqrt{10}} & =\frac{20 \sqrt{10}}{10} \\ & =2 \sqrt{10} \end{aligned}

Rationalising The Denominator - Examples 5 to 10

\begin{aligned} \frac{24}{\sqrt{15}} \times \frac{\sqrt{15}}{\sqrt{15}} & =\frac{24 \sqrt{15}}{15} \\ & =\frac{8 \sqrt{15}}{5} \end{aligned}

\begin{aligned} \frac{35}{4 \sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} & =\frac{35 \sqrt{5}}{20} \\ & =\frac{7 \sqrt{5}}{4} \end{aligned}

\frac{1}{9 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{18}

\begin{aligned} \frac{10-\sqrt{32}}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} & =\frac{\sqrt{2}(10-\sqrt{32})}{2} \\ & =\frac{10 \sqrt{2}-\sqrt{64}}{2} \\ & =\frac{10 \sqrt{2}-8}{2}=5 \sqrt{2}-4 \end{aligned}

\begin{aligned} \frac{\sqrt{12}+9}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} & =\frac{\sqrt{3}(\sqrt{12}+9)}{3} \\ & =\frac{\sqrt{36}+9 \sqrt{3}}{3} \\ & =\frac{6+9 \sqrt{3}}{3}=2+3 \sqrt{3} \end{aligned}

\frac{1^5}{(\sqrt{2})^5}=\frac{1}{4 \sqrt{2}} \quad \frac{1}{4 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{8}
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Rationalising The Denominator – Examples 11 to 17




 
\begin{aligned} 20 \times \sqrt{\frac{16}{5}} & =20 \times \frac{\sqrt{16}}{\sqrt{5}} \\ & =20 \times \frac{4}{\sqrt{5}} \\ & =\frac{80}{\sqrt{5}} \end{aligned}
 
\begin{aligned} & =\frac{80}{\sqrt{5}} \times \frac{\sqrt{5}}{\sqrt{5}} \\ & =\frac{80 \sqrt{5}}{5} \\ & =16 \sqrt{5} \end{aligned}
 
\begin{aligned} \frac{\sqrt{3}+\sqrt{5}}{\sqrt{2}} & \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{6}+\sqrt{10}}{2} \\ \frac{5}{\sqrt{10}} \times \frac{\sqrt{10}}{\sqrt{10}} & =\frac{5 \sqrt{10}}{10} \\ & =\frac{\sqrt{10}}{2} \end{aligned}
 
\begin{aligned} & \frac{\sqrt{6}+\sqrt{10}}{2}-\frac{\sqrt{10}}{2} \\ & =\frac{\sqrt{6}+\sqrt{10}-\sqrt{10}}{2} \\ & =\frac{\sqrt{6}}{2} \end{aligned}

\begin{aligned} \frac{1}{3-\sqrt{2}} \times \frac{3+\sqrt{2}}{3+\sqrt{2}} & =\frac{3+\sqrt{2}}{(3-\sqrt{2})(3+\sqrt{2})} \\ & =\frac{3+\sqrt{2}}{9+3 \sqrt{2}-3 \sqrt{2}-2} \\ & =\frac{3+\sqrt{2}}{7} \end{aligned}

\begin{aligned} \frac{7}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}} & =\frac{7(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})} \\ & =\frac{14-7 \sqrt{3}}{4-2 \sqrt{3}+2 \sqrt{3}-3} \\ & =\frac{14-7 \sqrt{3}}{1} \\ & =14-7 \sqrt{3} \end{aligned}

\begin{aligned} \frac{\sqrt{2}}{\sqrt{6}-2} \times \frac{\sqrt{6}+2}{\sqrt{6}+2} & =\frac{\sqrt{2}(\sqrt{6}+2)}{(\sqrt{6}-2)(\sqrt{6}+2)} \\ & =\frac{\sqrt{12}+2 \sqrt{2}}{6+2 \sqrt{6}-2 \sqrt{6}-4} \\ & =\frac{\sqrt{12}+2 \sqrt{2}}{2} \end{aligned}\begin{aligned} & =\frac{\sqrt{4} \times \sqrt{3}+2 \sqrt{2}}{2} \\ & =\frac{2 \sqrt{3}+2 \sqrt{2}}{2} \\ & =\sqrt{3}+\sqrt{2} \end{aligned}
 
 
\begin{aligned} \frac{4 \sqrt{5}}{3+\sqrt{10}} \times \frac{3-\sqrt{10}}{3-\sqrt{10}} & =\frac{4 \sqrt{5}(3-\sqrt{10})}{(3+\sqrt{10})(3-\sqrt{10})} \\ & =\frac{12 \sqrt{5}-4 \sqrt{50}}{9-3 \sqrt{10}+3 \sqrt{10}-10} \\ & =\frac{12 \sqrt{5}-4 \times \sqrt{25} \times \sqrt{2}}{-1} \\ & =20 \sqrt{2}-12 \sqrt{5} \end{aligned}
 
 
 
\begin{aligned} \frac{8-2 \sqrt{3}}{5-\sqrt{3}} \times \frac{5+\sqrt{3}}{5+\sqrt{3}} & =\frac{(8-2 \sqrt{3})(5+\sqrt{3})}{(5-\sqrt{3})(5+\sqrt{3})} \\ & =\frac{40+8 \sqrt{3}-10 \sqrt{3}-6}{25+5 \sqrt{3}-5 \sqrt{3}-3} \\ & =\frac{34-2 \sqrt{3}}{22} \\ & =\frac{17-\sqrt{3}}{11} \end{aligned}
 

Conclusion:
 Rationalising the denominator in GCSE Maths involves multiplying both the numerator and denominator by the conjugate of the denominator. This process eliminates square roots or other irrational values from the denominator, resulting in a rationalised fraction.

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