Question

C part with explanation

Answer 1

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

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

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

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

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