Question

if p=2-√3 and q=√3-√7 then find the value of (p^2+q^2)

Answer 1

Answer:

$17 - 2 \sqrt{3} (2 + \sqrt{7} )$

Step-by-step explanation:

$\huge\green{{Answer \: is ........}}$

$p = 2 - \sqrt{3} \\ q = \sqrt{3} - \sqrt{7} \\ {p}^{2} + {q}^{2} = {(2 - \sqrt{3})}^{2} + {( \sqrt{3} - \sqrt{7})}^{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 4 + 3 - 4 \sqrt{3} + 3 + 7 - 2 \sqrt{21} \\ \: \: \: \: \: \: \: \: \: \: = 17 - 2 \sqrt{3} (2 + \sqrt{7} )$

Answer 2

Answer:

-25 -4√3

Step-by-step explanation:

Given: p=2-√3 and q=√3-√7

by substituting the values of p and q in

(p^2+q^2) = (2-√3)^2 + (√3-√7)^2

                = (4 - 4√3 + 3) + (3 - 2√3√7 + 7)

                = 7 - 4√3 + 10 - 42

                = -25 -4√3