Answer:

– 1: Find the required values.

Given, x+

x

1

=4........(1)

We know,

⇒(x+

x

1

)

2

=x

2

+

x

2

1

+2

⇒(4)

2

=x

2

+

x

2

1

+2 [From (1)]

⇒16=x

2

+

x

2

1

+2

⇒x

2

+

x

2

1

=16−2

⇒x

2

+

x

2

1

=14........(2)

Step – 2: Find value of the square of x

2

+

x

2

1

Again,

⇒(x

2

+

x

2

1

)

2

=(x

2

)

2

+

(x

2

)

2

1

+2

⇒14

2

=x

4

+

x

4

1

+2 [From (2)]

⇒x

4

+

x

4

1

=196−2=194

Hence, the required values are (i) x

2

+

x

2

1

=1

Verified answer

Questions:-

[tex] \rm \: If \: \: x + \dfrac{1}{x} = 5[/tex]

Then find the value of

[tex] \rm \: 1. \: {x}^{2} + \dfrac{1}{ {x}^{2} } [/tex]

[tex] \rm \: 2. \: {x}^{4} + \dfrac{1}{ {x}^{4} } [/tex]

[tex] \rm \: 3.x - \dfrac{1}{x} [/tex]

ANSWER:-

[tex] \rm \: 1. \: {x}^{2} + \dfrac{1}{ {x}^{2} } = 23[/tex]

[tex] \rm \: 2. \: {x}^{4} + \dfrac{1}{ {x}^{4} } = 527[/tex]

[tex] \rm \: 3.x - \dfrac{1}{x} = \sqrt{21} [/tex]

Step-by-step explanation:

1))))

[tex] \rm \implies\: x + \dfrac{1}{x} = 5[/tex]

Squaring both sides

[tex] \rm \implies\bigg(x + \dfrac{1}{x} \bigg)^{2} =( 5)^{2} [/tex]

[tex] \rm \implies(x )^{2} + \bigg( \dfrac{1}{ {x} } \bigg) ^{2} + 2.x. \dfrac{1}{x} =25[/tex]

[tex] \rm \implies x ^{2} + \dfrac{1}{ { {x}^{2} } } + 2. \cancel{x}. \dfrac{1}{ \cancel x} =25[/tex]

[tex] \rm \implies x ^{2} + \dfrac{1}{ { {x}^{2} } } + 2=25[/tex]

[tex] \rm \implies x ^{2} + \dfrac{1}{ { {x}^{2} } } =25 - 2[/tex]

[tex] \rm \implies x ^{2} + \dfrac{1}{ { {x}^{2} } } =23[/tex]

2)))))

we got

[tex] \rm x ^{2} + \dfrac{1}{ { {x}^{2} } } =23[/tex]

Squaring both sides,

[tex] \rm \implies \bigg(x ^{2} + \dfrac{1}{ { {x}^{2} } } \bigg)^{2} =(23)^{2} [/tex]

[tex] \rm \implies (x ^{2})^{2} + \bigg(\dfrac{1}{ { {x}^{2} } } \bigg) ^{2} + 2. {x}^{2}. \dfrac{1}{ {x}^{2} } =529[/tex]

[tex] \rm \implies x ^{4} + \dfrac{1} { {x}^{4} } + 2. \cancel{ {x}^{2}}. \dfrac{1}{ \cancel{ {x}^{2} } } =529[/tex]

[tex] \rm \implies x ^{4} + \dfrac{1} { {x}^{4} } + 2 =529[/tex]

[tex] \rm \implies x ^{4} + \dfrac{1} { {x}^{4} } =529 - 2[/tex]

[tex] \rm \implies x ^{4} + \dfrac{1} { {x}^{4} } =527[/tex]

3)))))))

We know

[tex] \rm \bigg(x - \dfrac{1}{x} \bigg)^{2} = {x}^{2} + \dfrac{1}{ {x}^{2} } - 2.x. \dfrac{1}{x} [/tex]

[tex] \implies \rm \bigg(x - \dfrac{1}{x} \bigg)^{2} = {x}^{2} + \dfrac{1}{ {x}^{2} } - 2[/tex]

From Question 1))) We got

[tex] \rm x ^{2} + \dfrac{1}{ { {x}^{2} } } =23[/tex]

put it,

[tex] \implies \rm \bigg(x - \dfrac{1}{x} \bigg)^{2} = 23 - 2[/tex]

[tex] \implies \rm \bigg(x - \dfrac{1}{x} \bigg)^{2} = 21[/tex]

[tex] \implies \rm x - \dfrac{1}{x} = \sqrt{21} [/tex]

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[tex]\begin{gathered}\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: identities}}}} \\ \\ \bigstar \: \bf{ {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bigstar \: \bf{ {(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2} }\: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bigstar \: \bf{ {x}^{2} - {y}^{2} = (x + y)(x - y)}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bigstar \: \bf{ {(x + y)}^{2} - {(x - y)}^{2} = 4xy} \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \bigstar \: \bf{ {(x + y)}^{2} + {(x - y)}^{2} = 2( {x}^{2} + {y}^{2})}\:\\ \\ \bigstar \: \bf{ {(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)}\:\\ \\ \bigstar \: \bf{ {(x - y)}^{3} = {x}^{3} - {y}^{3} - 3xy(x - y) }\:\\ \\ \bigstar \: \bf{ {x}^{3} + {y}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )}\: \end{array} }}\end{gathered}\end{gathered}\end{gathered}\end{gathered}[/tex]