Question

if p= 3 minus root 5 upon 3 + root 5 and q is equal to 3 + root 5 upon 3 minus root root 5 find the value of p square + q square

Answer 1

Answer:

188

Step-by-step explanation:

p=3-√5/3+√5

q=3+√5/3-√5

in p,

multiplying numerator and denominator by 3-√5

=3-√5/3+√5*3-√5/3-√5

=(3-√5)∧2/3-5

=9-6√5+5/-2

=14-6√5/-2

=2(7-3√5)/-2

=-7+3√5

now in q,

multiplying numerator and denominator by 3+√5/3+√5

=3+√5/3-√5*3+√5/3+√5

=(3+√5)^2/-2

=9+6√5+5/-2

=14+6√5/-2

=2(7+3√5)/-2

=-7-3√5

now

p^2+q^2=(-7+3√5)^2+(-7-3√5)^2

=49+42√5+45+49-42√5+45

=98+90

=188

Answer 2

Answer:

Value f p² + q² is 47.

Step-by-step explanation:

Given:

$p=\frac{3-\sqrt{5}}{3+\sqrt{5}}\:\:and\:\:q=\frac{3+\sqrt{5}}{3-\sqrt{5}}$

To find: Value of p² + q²

Consider,

$p=\frac{3-\sqrt{5}}{3+\sqrt{5}}$

$=\frac{3-\sqrt{5}}{3+\sqrt{5}}\times\frac{3-\sqrt{5}}{3-\sqrt{5}}$

$=\frac{(3-\sqrt{5})(3-\sqrt{5})}{(3+\sqrt{5})(3-\sqrt{5})}$

$=\frac{(3)^2+(\sqrt{5})^2-2(3)(\sqrt{5})}{(3)^2-(\sqrt{5})^2}$

$=\frac{9+5-6\sqrt{5}}{9-5}$

$=\frac{14-6\sqrt{5}}{4}$

$=\frac{7-3\sqrt{5}}{2}$

$\implies p^2=(\frac{7-3\sqrt{5}}{2})^2=\frac{(7-3\sqrt{5})^2}{2^2}=\frac{49+45-42\sqrt{5}}{4}=\frac{94-42\sqrt{5}}{4}=\frac{47-21\sqrt{5}}{2}$

Now, Consider

$q=\frac{3+\sqrt{5}}{3-\sqrt{5}}$

$=\frac{3+\sqrt{5}}{3-\sqrt{5}}\times\frac{3+\sqrt{5}}{3+\sqrt{5}}$

$=\frac{(3+\sqrt{5})(3+\sqrt{5})}{(3-\sqrt{5})(3+\sqrt{5})}$

$=\frac{(3)^2+(\sqrt{5})^2+2(3)(\sqrt{5})}{(3)^2-(\sqrt{5})^2}$

$=\frac{9+5+6\sqrt{5}}{9-5}$

$=\frac{14+6\sqrt{5}}{4}$

$=\frac{7+3\sqrt{5}}{2}$

$\implies q^2=(\frac{7+3\sqrt{5}}{2})^2=\frac{(7+3\sqrt{5})^2}{2^2}=\frac{49+45+42\sqrt{5}}{4}=\frac{94+42\sqrt{5}}{4}=\frac{47+21\sqrt{5}}{2}$

So,

p² + q² = $\frac{47-21\sqrt{5}}{2}+\frac{47+21\sqrt{5}}{2}=\frac{47-21\sqrt{5}+47+21\sqrt{5}}{2}=\frac{2\times47}{2}=47$

Therefore, Value of p² + q² is 47.