Question

Q11. If p=1+√3, then find the value of
(i) p2 + 1/p2
(ii) p4 + 1/p4

Answer 1

Answer:

hello mate here is your answer

Step-by-step explanation:

hope it's help for you

Answer 2

Answer:

(i) p²+ $\frac{1}{p^2}$ = $\frac{1}{2}$(10 +3√3)

(ii) p⁴ + $\frac{1}{p^4}$ = $\frac{1}{4}$(119 + 60√3 )

Step-by-step explanation:

Given,

p = 1+√3

To find,

(i) p² + $\frac{1}{p^2}$

(ii) p⁴ + $\frac{1}{p^4}$

Recall the formula

(a+b)² = a² + 2ab + b² ---------------(1)

Solution

Since p = 1+√3, we have

(i)  $\frac{1}{p}$ = $\frac{1}{1+\sqrt{3} }$

= $\frac{1}{1+\sqrt{3} } X \frac{1 -\sqrt{3} }{1-\sqrt{3} }$

= $\frac{1 -\sqrt{3} }{1-3 }$

= $\frac{1 -\sqrt{3} }{-2}$

= $\frac{\sqrt{3} - 1 }{2}$

Applying a = p and b = $\frac{1}{p}$ in the identity (1), we get

(p + $\frac{1}{p}$)² = p²+ $\frac{1}{p^2}$ + 2×p× $\frac{1}{p}$

$((1+\sqrt{3} ) + ( \frac{\sqrt{3} - 1 }{2} ))^2$= p²+ $\frac{1}{p^2}$ + 2

$(\frac{2 +2\sqrt{3} +\sqrt{3} - 1 }{2})^2$ = p²+ $\frac{1}{p^2}$ + 2

$(\frac{1 +3\sqrt{3} }{2})^2$ = p²+ $\frac{1}{p^2}$ + 2

$(\frac{1 +9X3 + 2X3\sqrt{3} }{2})^2$ = p²+ $\frac{1}{p^2}$ + 2

$\frac{28+ 6\sqrt{3} }{4} - 2$ = p²+ $\frac{1}{p^2}$

p²+ $\frac{1}{p^2}$ = $\frac{28+ 6\sqrt{3} -8 }{4}$

=  $\frac{20+ 6\sqrt{3} }{4}$

= $\frac{10+ 3\sqrt{3} }{2}$

∴ p²+ $\frac{1}{p^2}$ =  $\frac{10+ 3\sqrt{3} }{2}$

(ii) We have p²+ $\frac{1}{p^2}$ =  $\frac{10+ 3\sqrt{3} }{2}$

( p²+ $\frac{1}{p^2}$ )²= ( $\frac{10+ 3\sqrt{3} }{2}$)²

p⁴ + $\frac{1}{p^4}$ + 2×p²× $\frac{1}{p^2}$ = $\frac{1}{4}$(10² + 3² × 3 + 2×10×3√3)

p⁴ + $\frac{1}{p^4}$ + 2 =  $\frac{1}{4}$(100 + 27 + 60√3)

=  $\frac{1}{4}$(127 + 60√3)

p⁴ + $\frac{1}{p^4}$ =   $\frac{1}{4}$(127 + 60√3) - 2

=  $\frac{1}{4}$(127 + 60√3 - 8)

=  $\frac{1}{4}$(119 + 60√3 )

p⁴ + $\frac{1}{p^4}$ = $\frac{1}{4}$(119 + 60√3 )

(i)p²+ $\frac{1}{p^2}$ = $\frac{1}{2}$(10 +3√3)

(ii) p⁴ + $\frac{1}{p^4}$ = $\frac{1}{4}$(119 + 60√3 )

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