Question

If (x+1/x)^2 = 96 then what is x^2 + 1/x^2 = ?

Answer 1

Answer: 94

Work Shown:

`\left(\text{x} + \frac{1}{\text{x}}\right)^2 = 96\\\\\left(\text{x}\right)^2 + 2*\text{x}*\frac{1}{\text{x}} + \left(\frac{1}{\text{x}}\right)^2 = 96\\\\\text{x}^2 + 2 + \frac{1}{\text{x}^2} = 96\\\\\text{x}^2 + \frac{1}{\text{x}^2} = 96 - 2\\\\\text{x}^2 + \frac{1}{\text{x}^2} = 94\\\\`

Step-by-step explanation:

I used the rule (A+B)^2 = A^2+2AB+B^2 on the second step. Conveniently the x terms cancel when computing the 2AB portion. This yields a constant that can be moved over to the other side to isolate the
`\text{x}^2+\frac{1}{\text{x}^2}` portion.

A longer method would be to solve for x to get four approximate roots. Those roots are

x = -0.10315

x = 0.10315

x = -9.6948

x = 9.6948

Then plug one of those roots into
`\text{x}^2 + \frac{1}{\text{x}^2}`

Let's try x = -0.10315

`\text{x}^2 + \frac{1}{\text{x}^2} \approx (-0.10315)^2 + (1)/((-0.10315)^2) \approx 93.996287` which is really close to 94. There's rounding error going on. Use more decimal digits in each root to get a more accurate value for
`\text{x}^2 + \frac{1}{\text{x}^2}`

I'll let you try the other roots.

I recommend using the first method since it's faster and more direct. Also, there isn't any rounding error to worry about.