Question

If p is equals to root 7 minus root 5 by root 7 + root 5 and q is equals to root 7 + root 5 by root 7 + root 5 find the value of p square minus q squared

Answer 1

Hey friend,
Here is the answer you were looking for:

$p = \frac{ \sqrt{7} - \sqrt{5} }{ \sqrt{7} + \sqrt{5} } \\ \\ on \: rationalizing \: the \: denominator \: we \: get \\ \\ p = \frac{ \sqrt{7} - \sqrt{5} }{ \sqrt{7} + \sqrt{5} } \times \frac{ \sqrt{7} - \sqrt{5} }{ \sqrt{7} - \sqrt{5} } \\ \\ using \: the \: identity \\ {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ p = \frac{ {( \sqrt{7}) }^{2} + {( \sqrt{5} )}^{2} - 2( \sqrt{7})( \sqrt{5}) }{ {( \sqrt{7} )}^{2} - {( \sqrt{5} )}^{2} } \\ \\ p = \frac{7 + 5 - 2 \sqrt{35} }{7 - 5} \\ \\ p = \frac{12 - 2 \sqrt{35} }{2} \\ \\ p = 6 - \sqrt{35} \\ \\ q = \frac{ \sqrt{7} + \sqrt{5} }{ \sqrt{7} - \sqrt{5} } \\ \\ on \: rationalizing \: the \: denominator \: we \: get \\ \\ q = \frac{ \sqrt{7} + \sqrt{5} }{ \sqrt{7} - \sqrt{5} } \times \frac{ \sqrt{7} + \sqrt{5} }{ \sqrt{7} + \sqrt{5} } \\ \\ using \: the \: identity \\ {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ q = \frac{ {( \sqrt{7}) }^{2} + {( \sqrt{5} )}^{2} + 2( \sqrt{7})( \sqrt{5}) }{ {( \sqrt{7} )}^{2} - {( \sqrt{5} )}^{2} } \\ \\ q = \frac{7 + 5 + 2 \sqrt{35} }{7 - 5} \\ \\ q = \frac{12 + 2 \sqrt{35} }{2} \\ \\ q = 6 + \sqrt{35} \\ \\ now \\ {p}^{2} - {q}^{2} \\ \\ = {(6 - \sqrt{35}) }^{2} - {(6 + \sqrt{35}) }^{2} \\ \\ = ( {(6)}^{2} + {( \sqrt{35}) }^{2} - 2(6)( \sqrt{35} )) - ( {(6)}^{2} + {( \sqrt{35}) }^{2} + 2(6)( \sqrt{35} )) \\ \\ = (36 + 35 - 12 \sqrt{35} ) - (36 + 35 + 12 \sqrt{35} ) \\ \\ = 71 - 12 \sqrt{35} - 71 - 12 \sqrt{35} \\ \\ = - 12 \sqrt{35} - 12 \sqrt{35} \\ \\ = - 24 \sqrt{35}$

Hope this helps!!

If you have any doubt regarding to my answer, feel free to ask in the comment section or inbox me if needed.

@Mahak24

Thanks...
☺☺

Answer 2

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

$\large \bold{@WildCat7083}$

Things to know before solving this question,

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

[tex\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}}[/tex]

[tex\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}} \\ \\ \sf \dashrightarrow\:\dfrac{54 + 14\sqrt{5}}{44} - \dfrac{54 - 14\sqrt{5}}{44} = p - 7\sqrt{5}q[/tex]

Calculating value of p and q,

$\sf \dashrightarrow\:\dfrac{54 + 14\sqrt{5} - (54 - 14\sqrt{5})}{44} = p - 7\sqrt{5}q \\ \\ \sf\dashrightarrow\:\dfrac{54 + 14\sqrt{5} - 54 + 14\sqrt{5}}{44} = p - 7\sqrt{5}q \\ \\ \sf \dashrightarrow\:\dfrac{54 - 54 + 14\sqrt{5} + 14\sqrt{5}}{44} = p - 7\sqrt{5}q \\ \\ \sf \dashrightarrow\:\dfrac{(14 + 14)\sqrt{5}}{44} = p - 7\sqrt{5}q \\ \\ \sf \dashrightarrow\:\dfrac{28\sqrt{5}}{44} = p - 7\sqrt{5}q \\ \\ \sf \dashrightarrow\:\dfrac{\cancel{28}\sqrt{5}}{\cancel{44}} = p - 7\sqrt{5}q \\ \\ \sf \dashrightarrow\:\dfrac{7\sqrt{5}}{11} = p - 7\sqrt{5}q \\ \\ \dashrightarrow\:\sf \red{p = 0}\:and\:\red{q = \dfrac{-1}{11}} \\ \\ \therefore\:{\underline{\sf{Value\:of\:p\:and\:q\:=\:\bf{0}\:\sf{and}\:\bf{\dfrac{-1}{11}}}}}$