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:  The answer is 47.

Step-by-step explanation:  We are given that

$p=\dfrac{3-\sqrt5}{3+\sqrt5},~~q=\dfrac{3+\sqrt5}{3-\sqrt5}.$

We are to find the value of p² + q².

We have

$p^2+q^2\\\\=\left(\dfrac{3-\sqrt5}{3+\sqrt5}\right)^2+\left(\dfrac{3+\sqrt5}{3-\sqrt5}\right)^2\\\\\\=\dfrac{(3-\sqrt5)^4+(3+\sqrt5)^4}{(9-5)^2}\\\\\\=\dfrac{(14-6\sqrt5)^2+(14+6\sqrt5)^2}{16}\\\\\\=\dfrac{196-168\sqrt5+180+196+168\sqrt5+180}{16}\\\\\\=\dfrac{752}{16}\\\\=47.$

Thus, the required value is 47.

Answer 2

Given that,

$p=\dfrac{3-\sqrt5}{3+\sqrt5},~~q=\dfrac{3+\sqrt5}{3-\sqrt5}$

To find,

The value of $p^2+q^2$.

Solution,

We are given the values of p and q. So

$p^2+q^2=\left(\dfrac{3-\sqrt5}{3+\sqrt5}\right)^2+\left(\dfrac{3+\sqrt5}{3-\sqrt5}\right)^2\\\\\\=\dfrac{(3-\sqrt5)^4+(3+\sqrt5)^4}{(9-5)^2}\\\\\\=\dfrac{(14-6\sqrt5)^2+(14+6\sqrt5)^2}{16}\\\\\\=\dfrac{196-168\sqrt5+180+196+168\sqrt5+180}{16}\\\\\\=\dfrac{752}{16}\\\\=47.$

So, we have calculated the above value and we get the answer as 47.