Question

The roots of the equations 3(x^2+1/x^2)-16(x+1/x)+26=0​

Answer 1

3(x^2 +1/x^2) -16(x-1/x) + 26 = 0

The answers can not be -1, 3 or 1/3.

Pls. try substituting any of them and you will see that they do not satisfy the equation.

Let us assume x-1/x = y

Squaring both the sides, we get: (x-1/x)^2 = y^2

or, x^2 + 1/x^2 -2 = y^2

or, y^2 + 2 = x^2 + 1/x^2

substituting these values in the equation,we get

3(y^2 + 2) -16y + 26 = 0

or 3y^2 - 16y +32 = 0

solving this quadratic equation, we get: y = (16 +- sqrt(16^2 - 4*3*32))/(2*3)

or, y = (16 +- 8sqrt 2 i)/6 = (8+-4sqrt2 i)/3

now, y = x-1/x = (8+-sqrt(2)i)/3

This will be another quadratic equation set.

for example, let us take the first root of y, x-1/x = (8+sqrt(2)i)/3

on simplification we get: 3(x^1-1) = x(8+sqrt2 i)

i.e., 3x^2 -(8+isqrt2 )x -3 = 0

we can solve this to get some values of x and similarly try the other root of y for more values of x.

As you can see that -1,3 and 1/3 are not the solutions. Pls. check the equation again.