Question

If P is equal to 5 minus 2 root 6 then find p square + 1 by p square

Answer 1

Answer:

14 - 8√6 / 25 - 8√6

Step-by-step explanation:

Given :- P = 5 - 2√6..........(i)

To find :- P² + 1 / P²

Proof :- Substitute the value of p

(5 - 2√6)² + 1 / (5 - 2√6)²

(5)² - 2×2×2√6 - (2√6)² + 1 / (5)² - 2×2×2√6

25 - 8√6 - 12 + 1 / 25 - 8√6

25 - 12 + 1 - 8√6 / 25 - 8√6

13 + 1 - 8√6 / 25 - 8√6

14 - 8√6 / 25 - 8√6

I hope it helps.....

Answer 2

$p^{2} +\dfrac{1}{p^2}$  = 98

Step-by-step explanation:

We have,

P = $5-2\sqrt{6}$

To find, the value of $p^{2} +\dfrac{1}{p^2}$ = ?

∴ P = $5-2\sqrt{6}$

∴ $\dfrac{1}{p}=\dfrac{1}{5-2\sqrt{6}}$

Rationalising numerator and denominator, we get

$\dfrac{1}{p}=\dfrac{1}{5-2\sqrt{6}}\times \dfrac{5+2\sqrt{6}}{5+2\sqrt{6}}$

⇒ $\dfrac{1}{p}=\dfrac{5+2\sqrt{6}}{5^2-(2\sqrt{6})^2}$

⇒ $\dfrac{1}{p}=\dfrac{5+2\sqrt{6}}{25-24}$

⇒ $\dfrac{1}{p}=5+2\sqrt{6}$

∴ $p^{2} +\dfrac{1}{p^2}$

= $(5-2\sqrt{6})^2+(5+2\sqrt{6})^2$

Using the algebraic identity,

$(a+b)^{2} +(a-b)^{2}=2(a^2+b^2)$

$= 2[5^2+(2\sqrt{6)^2]$

= 2[25 + 4(6)]

= 2(25 + 24)

= 2(49)

= 98

∴ $p^{2} +\dfrac{1}{p^2}$  = 98