Question

If p p=2-√3, find the value of p²+1/p²

Answer 1

Answer:

${p}^{2} + \frac{1}{ {p}^{2} } = 14$

Step-by-step explanation:

$p = 2 - \sqrt{3} \\ {p}^{2} + \frac{1}{ {p}^{2} } \\ = {(2 - \sqrt{3} )}^{2} + \frac{1}{ {(2 - \sqrt{3}) }^{2} } \\ = 4 + 3 - 4 \sqrt{3} + \frac{1}{4 + 3 - 4 \sqrt{3} } \\ = 7 - 4 \sqrt{3} + \frac{1}{7 - 4 \sqrt{3} } \\ = \frac{ {(7 - 4 \sqrt{3} )}^{2} + 1 }{7 - 4 \sqrt{3} } \\ = \frac{49 + 48 - 56 \sqrt{3} + 1 }{7 - 4 \sqrt{3} } \\ = \frac{98 - 56 \sqrt{3} }{7 - 4 \sqrt{3} } \\ = \frac{14(7 - 4 \sqrt{3}) }{(7 - 4 \sqrt{3}) } \\ = 14$

Answer 2

GIVEN :–

• Value of p = 2-√3

TO FIND :–

• Value of p²+1/p² = ?

SOLUTION :–

$\\ \bf \implies {p}^{2} + \dfrac{1}{{p}^{2}} =?$

$\\ \bf \implies {p}^{2} + \dfrac{1}{{p}^{2}} = {p}^{2} + \dfrac{1}{{p}^{2}} + 2 - 2$

$\\ \bf \implies {p}^{2} + \dfrac{1}{{p}^{2}} = {p}^{2} + \dfrac{1}{{p}^{2}} + 2(p) \left( \dfrac{1}{p} \right)- 2$

$\\ \bf \implies {p}^{2} + \dfrac{1}{{p}^{2}} =\left[p + \dfrac{1}{p} \right]^{2} - 2$

$\\ \bf \implies {p}^{2} + \dfrac{1}{{p}^{2}} =\left[(2 - \sqrt{3})+ \dfrac{1}{(2 - \sqrt{3})} \right]^{2} - 2$

$\\ \bf \implies {p}^{2} + \dfrac{1}{{p}^{2}} =\left[(2 - \sqrt{3})+ \dfrac{1}{(2 - \sqrt{3})} \times \dfrac{(2 + \sqrt{3}) }{(2 + \sqrt{3})} \right]^{2} - 2$

$\\ \bf \implies {p}^{2} + \dfrac{1}{{p}^{2}} =\left[(2 - \sqrt{3})+\dfrac{2 + \sqrt{3}}{4 -3} \right]^{2} - 2$

$\\ \bf \implies {p}^{2} + \dfrac{1}{{p}^{2}} =\left[(2 - \sqrt{3})+(2 + \sqrt{3})\right]^{2} - 2$

$\\ \bf \implies {p}^{2} + \dfrac{1}{{p}^{2}} =(2 + 2)^{2} - 2$

$\\ \bf \implies {p}^{2} + \dfrac{1}{{p}^{2}} =(4)^{2} - 2$

$\\ \bf \implies {p}^{2} + \dfrac{1}{{p}^{2}} =16- 2$

$\\\implies \large{ \boxed{ \bf{p}^{2} + \dfrac{1}{{p}^{2}} =14}}$