Question

If
$x + \frac{1}{x} = 4$
find the value of:
(i)
$(x - \frac{1}{x}$
(ii)
$( {x}^{2} + \frac{1}{ {x}^{2} }$
(iii)
$( {x}^{4} + \frac{1}{ {x}^{4} } )$
Can anybody explain me this answer Step-by-step? It is from the Chapter Algebraic Identities. It's urgent. The best answer will be marked as the Brainliest.

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Answer 1

Given:

$x + \frac{1}{x} = 4$

To find:

The values of:

$i) \: x - \frac{1}{x} \\ \\ ii) {x}^{2} + \frac{1}{ {x}^{2} } \\ \\ iii) {x}^{4} + \frac{1}{ {x}^{4} }$

Solution:

i)

We have

$x + \frac{1}{x} = 4$

So, we will be using identities.

For first part we will usethe following two identities.

(a+b)² = a² +b² +2ab and

(a-b)² = a² + b² -2ab

$squaring \: both \: sides \\ = > \left(x + \frac{1}{x} \right ) {}^{2} = (4) {}^{2} \\ \\ = > {x}^{2} + \left( \frac{1}{x} \right) {}^{2} + 2 \times \cancel{x} \times \frac{1}{ \cancel{x} } = 16 \\ \\ = > {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 16 \\ \\ = > {x}^{2} + \frac{1}{x {}^{2} } = 16 - 2 \\ \\ = > {x}^{2} + \frac{1}{ {x}^{2} } = 14 \: - - (1) \\ \\ now \\ \\ \left( x - \frac{1}{x} \right) {}^{2} = {x}^{2} + \frac{1}{ {x}^{2} } - 2 \times \cancel{x} \times \frac{1}{ \cancel{x}} \\ \\ from \: (1) \\ \\ = > \left(x - \frac{1}{x} \right) {}^{2} = 14 - 2 \\ \\ = > \left (x - \frac{1}{x} \right) {}^{2} = 12 \\ \\ = > x - \frac{1}{x} = \sqrt{12} \\ \\ = > x - \frac{1}{x} = \sqrt{4 \times 3} \\ \\ we \: know \: that \: \sqrt{4} = 2 \\ \\ = > x - \frac{1}{x} = 2 \sqrt{3}$

Hence, 2√3 is the value.

__________________________

ii)

The above part has the answer of this part. The equation (1) is the answer.

We will use [ (a+b)² = a²+b²+2ab ]

$squaring \: both \: sides \\ = > \left(x + \frac{1}{x} \right ) {}^{2} = (4) {}^{2} \\ \\ = > {x}^{2} + \left( \frac{1}{x} \right) {}^{2} + 2 \times \cancel{x} \times \frac{1}{ \cancel{x} } = 16 \\ \\ = > {x}^{2} + \frac{1}{ {x}^{2} } + 2 = 16 \\ \\ = > {x}^{2} + \frac{1}{x {}^{2} } = 16 - 2 \\ \\ = > {x}^{2} + \frac{1}{ {x}^{2} } = 14$

Hence, 14 is the value.

______________________________

iii)

Again we will use,

(a+b)²= a² +b² +2ab

Here a= x² and b= 1/x²

$squaring \: both \: sides \\ = > \left( {x}^{2} + \frac{1}{ {x}^{2} } \right) {}^{2} = (14) {}^{2} \\ \\ \small = > ( {x}^{2}) {}^{2} + \left( \frac{1}{x {}^{2} } \right) {}^{2} + 2 \times \cancel{ {x}^{2} } \times \frac{1}{ \cancel{ {x}^{2} }} = 196 \\ \\ = > {x}^{4} + \frac{1}{ {x}^{4} } + 2 = 196 \\ \\ = > {x}^{4} + \frac{1}{ {x}^{4} } = 196 - 2 \\ \\ = > {x}^{4} + \frac{1}{ {x}^{4} } = 194$

Hence, 194 is the value.

Hope it helps you.

Answer 2

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