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If \( \Large \frac{\sqrt{7}-2}{\sqrt{7}+2}=a\sqrt{7}+b \), then the value of a is:
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A) \( \Large \frac{11}{3} \) |
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B) \( \Large -\frac{4}{3} \) |
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C) \( \Large \frac{4}{3} \) |
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D) \( \Large \frac{-4\sqrt{7}}{3} \) |
Correct Answer:
| B) \( \Large -\frac{4}{3} \) |
Description for Correct answer:
\( \Large \frac{^{\sqrt{7}-2}}{\sqrt{7}+2}=a\sqrt{7}+b \)
L.H.S.=\( \Large \frac{\sqrt{7}-2}{\sqrt{7}+2} \times \frac{\sqrt{7}-2}{\sqrt{7}-2} \)
(Rationalisation)
\( \Large \frac{ \left(\sqrt{7}-2\right)^{2} }{ \left(\sqrt{7}\right)^{2}- \left(4\right) }=\frac{7+4-4\sqrt{7}}{7-4} \)
= \( \Large \frac{11-4\sqrt{7}}{3} \)
= \( \Large \frac{11}{3}-\frac{4}{3}\sqrt{7}=-\frac{4}{3}\sqrt{7}+\frac{11}{3} \)
=\( \Large a\sqrt{7}+b = R.H.S \)
\( \Large Compare\ the\ cofficients\ of\ \sqrt{7}\ and\ constant\ term \)
\( \Large a=-\frac{4}{3} \)
\( \Large b=\frac{11}{3} \)
\( \Large \frac{^{\sqrt{7}-2}}{\sqrt{7}+2}=a\sqrt{7}+b \)
L.H.S.=\( \Large \frac{\sqrt{7}-2}{\sqrt{7}+2} \times \frac{\sqrt{7}-2}{\sqrt{7}-2} \)
(Rationalisation)
\( \Large \frac{ \left(\sqrt{7}-2\right)^{2} }{ \left(\sqrt{7}\right)^{2}- \left(4\right) }=\frac{7+4-4\sqrt{7}}{7-4} \)
= \( \Large \frac{11-4\sqrt{7}}{3} \)
= \( \Large \frac{11}{3}-\frac{4}{3}\sqrt{7}=-\frac{4}{3}\sqrt{7}+\frac{11}{3} \)
=\( \Large a\sqrt{7}+b = R.H.S \)
\( \Large Compare\ the\ cofficients\ of\ \sqrt{7}\ and\ constant\ term \)
\( \Large a=-\frac{4}{3} \)
\( \Large b=\frac{11}{3} \)
Part of solved Elementary Mathematics questions and answers : >> Elementary Mathematics
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