Question

Thanks for all who helped me yesterday...Did my exam well...Want to check this one alone

Find the value of a and b if
$( \sqrt{7 } - 2) \div ( \sqrt{7} + 2) = a \sqrt{7} + b$
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Answer 1

Answer:

$\frac{\sqrt{7} -2}{\sqrt{7} +2} \\\\=\frac{(\sqrt{7} -2)^{2}}{(\sqrt{7} +2)({\sqrt{7} -2})} \\\\According\ to\ the\ formulae,\\\\(a-b)^{2} = a^{2} - 2ab - b^{2} \\(a+b)(a-b) = a^{2}-b^{2}\\\\=\frac{(\sqrt{7})^{2}-2*2\sqrt{7} -2^{2} }{(\sqrt{7})^{2}-2^{2} } \\\\=\frac{7-[tex]=>\frac{3-4\sqrt{7}}{3}= a\sqrt{7} +b\\\\=>1-\frac{4\sqrt{7}}{3}= a\sqrt{7} +b\\\\=> a=\frac{-4}{3} , b= 1$

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