Question

Rationalise the denominator 4 /2 +root 3 +root 7

Answer 1

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Answer 2

Answer:

The rationalized value of $\frac{4}{2+\sqrt{3}+\sqrt{7}}$ is $\bold{\frac{2 \sqrt{3}+3-\sqrt{21}}{3}}$

Step-by-step explanation:

To rationalize $\frac{4}{2+\sqrt{3}+\sqrt{7}}$, “multiply the numerator and denominator” by $\bold{(2+\sqrt{3})-\sqrt{7}}$

$\begin{array}{l}{\frac{4}{2+\sqrt{3}+\sqrt{7}} \times \frac{(2+\sqrt{3})-\sqrt{7}}{(2+\sqrt{3})-\sqrt{7}}} \\ {x^{2}-y^{2}=(x+y)(x-y)}\end{array}$

Hence the equation becomes, $\frac{4(2+\sqrt{3})-\sqrt{7}}{(2+\sqrt{3})^{2}-(\sqrt{7})^{2}}$

$\begin{array}{l}{=\frac{4(2+\sqrt{3})-\sqrt{7}}{4+3+4 \sqrt{3}-7}} \\ {=\frac{4(2+\sqrt{3})-\sqrt{7}}{7+4 \sqrt{3}-7}} \\ {=\frac{2+\sqrt{3}-\sqrt{7}}{\sqrt{3}} * \frac{\sqrt{3}}{\sqrt{3}}} \\ {=\frac{2 \sqrt{3}+3-\sqrt{21}}{3}} \\ {=\frac{2 \sqrt{3}+3-\sqrt{21}}{3}}\end{array}$