Question

find p and q if 7 + root 5 by 7 minus root 5 minus 7 minus root 5 by 7 + root 5 is equals to p - 7 root 5 q​

Answer 1

EXPLANATION.

$\implies \dfrac{7 + \sqrt{5} }{7 - \sqrt{5} } - \dfrac{7 - \sqrt{5} }{7 + \sqrt{5} } = p - 7\sqrt{5} q$

As we know that,

Rationalizes the equation one by one, we get.

$\implies \dfrac{7 + \sqrt{5} }{7 - \sqrt{5} }$

$\implies \dfrac{7 + \sqrt{5} }{7 - \sqrt{5} } \times \dfrac{7 + \sqrt{5} }{7 + \sqrt{5} }$

$\implies \dfrac{(7 + \sqrt{5} )^{2} }{(7)^{2} - (\sqrt{5})^{2} }$

$\implies \dfrac{49 + 5 + 2 \times 7 \times \sqrt{5} }{49 - 5}$

$\implies \dfrac{54 + 14\sqrt{5} }{44}$

$\implies \dfrac{7 - \sqrt{5} }{7 + \sqrt{5} }$

$\implies \dfrac{7 - \sqrt{5} }{7 + \sqrt{5} } \times \dfrac{7 - \sqrt{5} }{7 - \sqrt{5} }$

$\implies \dfrac{(7 - \sqrt{5} )^{2} }{(7)^{2} - (\sqrt{5})^{2} }$

$\implies \dfrac{49 + 5 - 2 \times 7 \times \sqrt{5} }{49 - 5}$

$\implies \dfrac{54 - 14\sqrt{5} }{44}$

We can write equation as,

$\implies \dfrac{54 + 14\sqrt{5} }{44} - \dfrac{54 - 14\sqrt{5} }{44}$

$\implies \dfrac{54 + 14\sqrt{5} - 54 + 14\sqrt{5} }{44}$

$\implies \dfrac{14\sqrt{5} + 14\sqrt{5} }{44} = p - 7\sqrt{5} q$

$\implies \dfrac{28\sqrt{5} }{44} = p - 7\sqrt{5} q$

$\implies \dfrac{7\sqrt{5} }{11} = p - 7\sqrt{5} q$

$\implies p = 0 \ \ \ and \ \ \ q = \dfrac{-1}{11}$

Answer 2

$\large\underline{\sf{\red{Given}}}$

✭ $\sf \dfrac{7 + \sqrt{5}}{7 - \sqrt{5}} - \dfrac{7 - \sqrt{5}}{7 + \sqrt{5}} = p - 7\sqrt{5}q$

$\large\underline{\sf{\blue{To\:Find}}}$

✭ $\sf Value\:of\:p\:and\:q?$

$\large\underline{\sf{\green{Solution}}}$

Things to know before solving this question,

◈ $\underline{\boxed{\sf{(a + b)(a - b) = a^2 - b^2}}}$

◈ $\underline{\boxed{\sf{(a + b)^2 = a^2 + b^2 + 2ab}}}$

◈ $\underline{\boxed{\sf{(a - b)^2 = a^2 + b^2 - 2ab}}}$

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$\underline{\sf{\bigstar\:Rationalising\:the\:denominator\:of\:\frac{7 + \sqrt{5}}{7 - \sqrt{5}}\::}}$

➝ $\sf \dfrac{7 + \sqrt{5}}{7 - \sqrt{5}} \:\times\:\dfrac{7 + \sqrt{5}}{7 + \sqrt{5}}$

➝ $\sf \dfrac{(7 + \sqrt{5})^2}{(7)^2 - (\sqrt{5})^2}$

➝ $\sf \dfrac{(7)^2 + (\sqrt{5})^2 + 2(7)(\sqrt{5})}{49 - 5}$

➝ $\sf \dfrac{49 + 5 + 14\sqrt{5}}{44}$

➝ $\sf\pink{\dfrac{54 + 14\sqrt{5}}{44}}$

$\underline{\sf{\bigstar\:Rationalising\:the\:denominator\:of\:\frac{7 - \sqrt{5}}{7 + \sqrt{5}}\::}}$

➳ $\sf \dfrac{7 - \sqrt{5}}{7 + \sqrt{5}} \:\times\:\dfrac{7 - \sqrt{5}}{7 - \sqrt{5}}$

➳ $\sf \dfrac{(7 - \sqrt{5})^2}{(7)^2 - (\sqrt{5})^2}$

➳ $\sf \dfrac{(7)^2 + (\sqrt{5})^2 - 2(7)(\sqrt{5})}{49 - 5}$

➳ $\sf \dfrac{49 + 5 - 14\sqrt{5}}{44}$

➳ $\sf\pink{\dfrac{54 - 14\sqrt{5}}{44}}$

So, equation is,

$\sf \dashrightarrow\:\dfrac{54 + 14\sqrt{5}}{44} - \dfrac{54 - 14\sqrt{5}}{44} = p - 7\sqrt{5}q$

Calculating value of p and q,

$\sf \dashrightarrow\:\dfrac{54 + 14\sqrt{5} - (54 - 14\sqrt{5})}{44} = p - 7\sqrt{5}q$

$\sf \dashrightarrow\:\dfrac{54 + 14\sqrt{5} - 54 + 14\sqrt{5}}{44} = p - 7\sqrt{5}q$

$\sf \dashrightarrow\:\dfrac{54 - 54 + 14\sqrt{5} + 14\sqrt{5}}{44} = p - 7\sqrt{5}q$

$\sf \dashrightarrow\:\dfrac{(14 + 14)\sqrt{5}}{44} = p - 7\sqrt{5}q$

$\sf \dashrightarrow\:\dfrac{28\sqrt{5}}{44} = p - 7\sqrt{5}q$

$\sf \dashrightarrow\:\dfrac{\cancel{28}\sqrt{5}}{\cancel{44}} = p - 7\sqrt{5}q$

$\sf \dashrightarrow\:\dfrac{7\sqrt{5}}{11} = p - 7\sqrt{5}q$

$\dashrightarrow\:\sf \pink{p = 0}\:and\:\pink{q = \dfrac{-1}{11}}$

$\therefore\:{\underline{\sf{Value\:of\:p\:and\:q\:=\:\bf{0}\:\sf{and}\:\bf{\dfrac{-1}{11}}}}}$

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