y= x^2 + x^-2, find x when y =25.04

Answer 1

To find the value of x when y = 25.04, we can substitute y = 25.04 into the equation y = x^2 + x^-2:

25.04 = x^2 + x^-2

Now, we can rearrange the equation to get a quadratic equation in terms of x:

x^4 + 25.04x^2 + 1 = 0

This is a quadratic equation in terms of x^2. We can let z = x^2 to simplify the equation:

z^2 + 25.04z + 1 = 0

Now, we can use the quadratic formula to solve for z:

z = (-25.04 ± √(25.04^2 - 4*1*1)) / 2

z = (-25.04 ± √(627.2016 - 4)) / 2

z = (-25.04 ± √623.2016) / 2

z = (-25.04 ± 24.928) / 2

We get two possible values for z:

z1 = (-25.04 + 24.928) / 2 = -0.112
z2 = (-25.04 - 24.928) / 2 = -24.984

Since z = x^2, we can discard the negative value for z, as it would result in a complex value for x. Therefore, we take z1:

z = -0.112 = x^2

Taking the square root of both sides, we get:

x = ±√(-0.112)

This results in a complex value for x, so there is no real solution for x when y = 25.04 in the given equation y = x^2 + x^-2.