Question

This is very important question which will surely come. So, pls help!
_________________________
$\sf{If \: 3p + \dfrac{1}{3p} = 2 \sqrt{3},evaluate \: }\\\\\\\sf{(i)3p - \dfrac{1}{3p}}\\\\\sf{(ii)9 {p}^{2} + \frac{1}{9{p}^{2}}}\\\\ \sf{(iii)81 {p}^{4} + \frac{1}{81 {p}^{4}}.}$
Explain step by step.
Good Job!​

Answer 1

Hey Mate ! here are your Answers :

Given :

• 3p+1/3p = 2√3

To Find :

$\tt{ \bull 3p - \frac{1}{3p} }$

$\tt{ \bull {9p}^{2} + \frac{1}{ {9p}^{2} } }$

$\tt{ \bull {81p}^{4} + \frac{1}{ {81p}^{4} } }$

Solution :

1] As Given,

$\tt{ 3p + \frac{1}{3p} = 2 \sqrt{3} }$

By using the identity:

$\tt{ {(x - y)}^{2} + 4xy = {(x + y)}^{2} }$

We get, that :

$\tt{ \implies {(3p - \frac{1}{3p} )}^{2} +4 (3p \times \frac{1}{3p}) = {(3p + \frac{1}{3p} )}^{2}}$

$\tt{ {(3p - \frac{1}{3p}) }^{2} - 4 = {(2 \sqrt{3} )}^{2} }$

$\tt{ \implies {(3p - \frac{1}{3p} )}^{2} - 4 = 12}$

$\implies \tt{(3p - \frac{1}{3p} {)}^{2} = 12 - 4}$

$\tt{ \implies 3p - \frac{1}{3p} = \sqrt{8} }$

$\red{ \underline{ \boxed{ \tt{ \therefore \:3p - \frac{1}{3p} = 2 \sqrt{2}}}}}$

__________________________

2] As Given,

$\tt{3p + 1 \frac{1}{3p} = 2 \sqrt{3} }$

By using the identity:

$\tt{ {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}$

We get that :

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

$\tt{ \implies 9 {p}^{2} + \frac{1}{9 {p}^{2} } + 2 = 12}$

$\tt{ \implies \tt{9 {p}^{2} + \frac{1}{9 {p}^{2} } = 12 - 2}}$

$\green{ \underline{ \boxed{ \tt{ \therefore \: 9 {p}^{2} + \frac{1}{9 {p}^{2} } = 10}}}}$

__________________________

3] As we got,

$\green{ \underline{ \boxed{ \tt{ \therefore \: 9 {p}^{2} + \frac{1}{9 {p}^{2} } = 10}}}}$

Using the identity:

$\tt{ {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}$

We get that :

$\tt{ \implies {(9 {p}^{2} + \frac{1}{9 {p}^{2} }) }^{2} = {10}^{2} }$

$\tt{ \implies {(9 {p}^{2}) }^{2} + ({ \frac{1}{9 {p}^{2} } )}^{2} +2 = 100 }$

$\tt{ \implies 81 {p}^{4} + \frac{1}{81 {p}^{4} } = 100 - 2}$

$\blue{ \underline{ \boxed{ \tt{ \therefore \: 81 {p}^{4} + \frac{1}{81 {p}^{4} } = 98}}}}$

Best of luck for exams !

Answer 2

ey Mate ! here are your Answers :

Given :

3p+1/3p = 2√3

To Find :

\tt{ \bull 3p - \frac{1}{3p} }∙3p−3p1

\tt{ \bull {9p}^{2} + \frac{1}{ {9p}^{2} } }∙9p2+9p21

\tt{ \bull {81p}^{4} + \frac{1}{ {81p}^{4} } }∙81p4+81p41

Solution :

1] As Given,

\tt{ 3p + \frac{1}{3p} = 2 \sqrt{3} }3p+3p1=23

By using the identity:

\tt{ {(x - y)}^{2} + 4xy = {(x + y)}^{2} }(x−y)2+4xy=(x+y)2

We get, that :

\tt{ \implies {(3p - \frac{1}{3p} )}^{2} +4 (3p \times \frac{1}{3p}) = {(3p + \frac{1}{3p} )}^{2}}⟹(3p−3p1)2+4(3p×3p1)=(3p+3p1)2

\tt{ {(3p - \frac{1}{3p}) }^{2} - 4 = {(2 \sqrt{3} )}^{2} }(3p−3p1)2−4=(23)2

\tt{ \implies {(3p - \frac{1}{3p} )}^{2} - 4 = 12}⟹(3p−3p1)2−4=12

\implies \tt{(3p - \frac{1}{3p} {)}^{2} = 12 - 4}⟹(3p−3p1)2=12−4

\tt{ \implies 3p - \frac{1}{3p} = \sqrt{8} }⟹3p−3p1=8

\red{ \underline{ \boxed{ \tt{ \therefore \:3p - \frac{1}{3p} = 2 \sqrt{2}}}}}∴3p−3p1=22

__________________________

2] As Given,

\tt{3p + 1 \frac{1}{3p} = 2 \sqrt{3} }3p+13p1=23

By using the identity:

\tt{ {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy}(x+y)2=x2+y2+2xy

We get that :

\tt{ \implies {(3p + \frac{1}{3p} )}^{2} = (2 \sqrt{3} {)}^{2} }⟹(3p+3p1)2=(23)2

\tt{ \implies 9 {p}^{2} + \frac{1}{9 {p}^{2} } + 2 = 12}⟹9p2+9p21+2=12

\tt{ \implies \tt{9 {p}^{2} + \frac{1}{9 {p}^{2} } = 12 - 2}}⟹9p2+9p21=12−2

\green{ \underline{ \boxed{ \tt{ \therefore \: 9 {p}^{2} + \frac{1}{9 {p}^{2} } = 10}}}}∴9p2+9p21=10

__________________________

3] As we got,

\green{ \underline{ \boxed{ \tt{ \therefore \: 9 {p}^{2} + \frac{1}{9 {p}^{2} } = 10}}}}∴