Question

If p=3-√5/3+√5 and q=3+√5/3-√5 then find the value of p^2+q^2

Answer 1

Answer:

47

Step-by-step explanation:

p and q needs rationalization.

∴$p=\frac{(3-\sqrt{5})^2 }{3^2-(\sqrt{5})^2 } =\frac{14-6\sqrt{5} }{4} =\frac{7-3\sqrt{5} }{2}$

∴$q=\frac{(3+\sqrt{5})^2 }{3^2-(\sqrt{5})^2 } =\frac{14+6\sqrt{5} }{4} =\frac{7+3\sqrt{5} }{2}$

Thus,

$pq=\frac{7^2-(3\sqrt{5})^2 }{4} =\frac{4}{4} =1$ → (A)

$p+q=\frac{14}{2} =7$ → (B).

$p^2+q^2=(p+q)^2-2pq$ → (C)

∵On the other hand, the identity $p^2+2pq+q^2=(p+q)^2$ can be used

to obtain $p^2+q^2=(p+q)^2-2pq$.

∴Therefore, $p^2+q^2=49-2=47$.