Question

if p=4+√5/4-√5 and q=4-√5/4+ √5 find p^2 +q^2

Answer 1

$\huge\mathbb{\underline{\pink{SOLUTION:-}}}$

$\bold{Given:-}\begin{cases}\sf{p=\frac{4+\sqrt{5}}{4-\sqrt{5}}}\\ \\ \sf{q=\frac{4-\sqrt{5}}{4+\sqrt{5}}}\end{cases}$

• We have to find the value of
• $\tt p{}^{2}+q{}^{2}$

$\bf\underline{\underline{According\:to\: Question:-}}$

• 1st Rationalize the value of

$\sf\implies p=\frac{4+\sqrt{5}}{4-\sqrt{5}}\\ \\ \sf\implies p=\frac{4+\sqrt{5}}{4-\sqrt{5}}\times \frac{4+\sqrt{5}}{4+\sqrt{5}}\\ \\ \sf\implies p=\frac{(4+\sqrt{5}){}^{2}}{(4){}^{2}-(\sqrt{5}){}^{2}}\\ \\ \sf\implies p=\frac{(4){}^{2}+(\sqrt{5}){}^{2}+2\times 4\times \sqrt{5}}{16-5}\\ \\ \sf\implies p=\frac{16+5+8\sqrt{5}}{11}\\ \\ \sf\implies p=\frac{21+8\sqrt{5}}{11}$

• The value of p is
• $\sf \frac{21+8\sqrt{5}}{11}$

• Now the value of q

$\sf\implies q=\frac{4-\sqrt{5}}{4+\sqrt{5}}\\ \\ \sf\implies q=\frac{4-\sqrt{5}}{4+\sqrt{5}}\times \frac{4-\sqrt{5}}{4-\sqrt{5}}\\ \\ \sf\implies q=\frac{(4-\sqrt{5}){}^{2}}{(4){}^{2}-(\sqrt{5}){}^{2}}\\ \\ \sf\implies q=\frac{(4){}^{2}+(\sqrt{5}){}^{2}-2\times 4\times \sqrt{5}}{16-5}\\ \\ \sf\implies q=\frac{16+5-8\sqrt{5}}{11}\\ \\ \sf\implies q=\frac{21-8\sqrt{5}}{11}$

• The value of q is
• $\sf q=\frac{21-8\sqrt{5}}{11}$

$\sf The\:value\:of\:p{}^{2}\\ \\ \sf\implies [\frac {21+8\sqrt{5}}{11}]{}^{2}=\frac{(21){}^{2}+(8\sqrt{5}){}^{2}}{(11){}^{2}}\\ \\ \sf\implies p{}^{2}=\frac{441+64\times 5}{121}\\ \\ \sf\implies p{}^{2}=\frac{441+320}{121}=\frac{761}{121}$

$\bf\implies p{}^{2}=\frac{761}{121}$

$\sf The\:value\:of\:q{}^{2}\\ \\ \sf\implies [\frac {21-8\sqrt{5}}{11}]{}^{2}=\frac{(21){}^{2}-(8\sqrt{5}){}^{2}}{(11){}^{2}}\\ \\ \sf\implies q{}^{2}=\frac{441-64\times 5}{121}\\ \\ \sf\implies q{}^{2}=\frac{441-320}{121}=\cancel{\frac{121}{121}}=1$

$\bf\implies q{}^{2}=1$

$\boxed{\mathfrak{p{}^{2}=\frac{761}{121}}}$

$\boxed{\mathfrak{q{}^{2}=1}}$

$\boxed{\sf{Now\: p{}^{2}+q{}^{2}}}$

$\sf \implies p{}^{2}+q{}^{2}=\frac{761}{121}+1\\ \\ \sf\implies p{}^{2}+q{}^{2}=\frac{761+121}{121}\\ \\ \sf\implies p{}^{2}+q{}^{2}=\frac{882}{121}$

$\boxed{\mathfrak{\red{p{}^{2}+q{}^{2}=\frac{882}{121}}}}$

#BAL

#answerwithquality