Question

Find:-x+1/x if x to the power 4+1/x to the power 4 is 47

Answer 1

3.

Step-by-step explanation:

Given:

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

To find:

$x + \dfrac{1}{x}$

How can we solve ?

1. By Squaring and elimination of square.

2. Using Algebra Formula

3. Simplify and Answer.

Solution:

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

[Adding 2 in both sides]

${x}^{4} + \dfrac{1}{ {x}^{4} } + 2 = 47 + 2 \\ \\ {x}^{4} + \frac{1}{ {x}^{4} } + 2. {x}^{2} . \frac{1}{ {x}^{2} } = 49 \\ \\ {( {x}^{2} + \frac{1}{ {x}^{2} })}^{2} = {7}^{2}$

[Eliminating Square from both sides]

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

[Adding 2 in both sides]

${x}^{2} + \frac{1}{ {x}^{2} } + 2 = 7 + 2 \\ \\ {x}^{2} + \frac{1}{ {x}^{2} } + 2.x. \frac{1}{x} = 9 \\ \\ {(x + \frac{1}{x})}^{2} = {3}^{2}$

[Eliminating Square from both sides]

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

The required answer is 3.

Some Important Algebra Formula

${(x + y)}^{2}={x}^{2}+{y}^{2}+2xy\\ \\{(x - y)}^{2}={x}^{2}+{y}^{2}-2xy\\ \\{(x+y)}^{2}= (x - y) {}^{2}+4xy\\ \\{(x-y)}^{2}=(x+y){}^{2}-4xy\\ \\ (x + y)^{2}+(x-y)^{2}=2( {x}^{2}+{y}^{2} )\\ \\(x+y)^{2}- (x-y) {}^{2}=4xy$