# Solve- 3(x^2+1/x^2)-16(x-1/x)+26=0 Ans.-1,3,1/3 Please tell me how to solve it.

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For the given question: `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+i`sqrt2` )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.

Try and solve these quadratic equation to get the answer.

Hope this helps.

The solution you have provided are not of the equation 3(x^2+1/x^2)-16(x-1/x)+26=0. Rather they are the solutions of the equation 3(x^2+1/x^2)-16(x+1/x)+26=0. You have made a typo with the minus sign.

The equation 3(x^2+1/x^2)-16(x+1/x)+26=0 can be solved as follows.

Let x + 1/x = y, squaring both the sides x^2 + 1/x^2 + 2 = y^2 or x^2 + 1/x^2 = y^2 - 2

Substituting y in the original equation:

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

3y^2 - 6 - 16y + 26 = 0

3y^2 - 16y + 20 = 0

3y^2 - 6y - 10y + 20 = 0

3y(y - 2) - 10(y - 2) = 0

(3y - 10)(y - 2) = 0

y = 10/3, y = 2

y = x + 1/x

x + 1/x = 10/3

3x^2 - 10x + 3 = 0

3x^2 - 9x - x + 3 = 0

3x(x - 3)-1(x - 3) = 0

(3x - 1)(x - 3) = 0

x = 1/3, x = 3

x + 1/x = 2

x^2 - 2x + 1 = 0

(x - 1)^2 = 0

x = 1

The solution of the equation 3(x^2+1/x^2)-16(x+1/x)+26=0 are {1, 1/3, 3}