Math, asked by Ayushvaid, 1 year ago

C part with explanation

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Answers

Answered by nickkaushiknick
5

Answer:


Step-by-step explanation:

Here

x+\frac{1}{x}=3

Squaring both sides

(x+\frac{1}{x})^2=3^2

x^2 +\frac{1}{x^2}+2\times x \times \frac{1}{x} = 9

x^2 +\frac{1}{x^2}+2 = 9

x^2 +\frac{1}{x^2} = 7 ----------- ( i )

First Part:

We know that, (a - b)² = a² - 2ab + b²,  therefore

(x-\frac{1}{x})^2=x^2 +\frac{1}{x^2}-2 \times x \times \frac{1}{x}

(x-\frac{1}{x})^2=x^2+\frac{1}{x^2}-2

Putting value of [x² + 1/x²] from equation ( i )

(x-\frac{1}{x})^2= 7-2

(x-\frac{1}{x})^2= 5

(x-\frac{1}{x})= \sqrt 5

Second Part:

x+\frac{1}{x}=3

Squaring both sides

(x+\frac{1}{x})^2=3^2

x^2 +\frac{1}{x^2}+2\times x \times \frac{1}{x} = 9

x^2 +\frac{1}{x^2}+2 = 9

x^2 +\frac{1}{x^2} = 7

Third Part:

From equation ( i ) we get

x^2+\frac{1}{x^2}=7

Squaring both sides

(x^2+\frac{1}{x^2})^2=7^2

x^4+\frac{1}{x^4}+2(x^2)(\frac{1}{x^2})=49

x^4+\frac{1}{x^4}+2=49

x^4+\frac{1}{x^4}=47


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