Question

If (x-1/x) = 6, then find (x2 + 1/x2) and (x + 1/x)

I found (x2+ 1/x2) = 38

Now how to find (x + 1/x) ?

Answer 1

HEY HERE IS YOUR ANSWER ☺☺☺☺

($(x - \frac{1}{x}) = 6 \\ \\ {(x \ - \frac{1}{x}) }^{2} = {(6) }^{2} \\ \\ ( {x - \frac{1}{x}) }^{2} = {x}^{2} + \frac{1}{ {x}^{2} } - 2 \\ \\ {(6)}^{2} = {x}^{2} + \frac{1}{ {x}^{2} } - 2 \\ \\ 36 + 2 = {x}^{2} + \frac{1}{ {x}^{2} } \\ \\ 38 = {x}^{2} + \frac{1}{ {x}^{2} } \: \:$

38 = x^2 +1/x^2

$(x + \frac{1}{x} ) {}^{2} = {x}^{2} + \frac{1}{ {x}^{2} } + 2 \\ \\ (x + \frac{1}{x} ) {}^{2} = 38 + 2 \\ \\ (x + \frac{1}{x} ) {}^{2} = 40 \\ \\ (x + \frac{1}{x} ) = \sqrt{40} \\ \\ (x \ + \frac{1}{x} ) = \sqrt{2 \times 2 \times 2 \times 5 } \\ \\ (x + \frac{1}{x} ) = 2 \sqrt{10}$

HOPE IT HELPS ☺☺☺☺☺
PLEASE MARK AS BRAINLIEST
#AADI 93

Answer 2

Answer:

If $x -\frac{1}{x}=6$             .....[1]

then:

find $x^2+\frac{1}{x^2}$

Squaring equation [1] both sides we get;

$(x -\frac{1}{x})^2=6^2$

Using identity rule:

$(a-b)^2 =a^2+b^2-2ab$

then;

$x^2+\frac{1}{x^2}-2x \cdot \frac{1}{x}= 36$

Simplify:

$x^2+\frac{1}{x^2}-2 =36$

Add 2 to both sides we get;

$x^2+\frac{1}{x^2}=38$

Now, find  $x+\frac{1}{x}$

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

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

Substitute the value of $x^2+\frac{1}{x^2}=38$ in above equation we have:

$(x+\frac{1}{x})=38+2$

Simplify:

$(x+\frac{1}{x})= 40$

Taking square root both sides we have;

$x+\frac{1}{x} = \sqrt{40}=2\sqrt{10}$

Therefore, the values of $x^2+\frac{1}{x^2}=38$ and  $x+\frac{1}{x} = 2\sqrt{10}$