Question

Please solve this equation

Answer 1

Answer:

The result for the given expression will be:

$x=4\\x=\dfrac{1}{4}\\x=\dfrac{-9+\sqrt {65}}{4}\\x=\dfrac{-9-\sqrt {65}}{4}$

Step-by-step explanation:

The given equation is:

$8(x-\dfrac{1}{x})^2+2(x+\dfrac{1}{x})=121$

Expanding the function in the form of:

$(a-b)^2=a^2-2ab+b^2$.

$8(x^2-2x\cdot \dfrac{1}{x}+\dfrac{1}{x^2})+2(x+\dfrac{1}{x})=121\\8(x^2-2+\dfrac{1}{x^2})+2(x+\dfrac{1}{x})=121\\8x^2-16+\dfrac{8}{x^2}+2x+\dfrac{2}{x}=121\\8x^2+\dfrac{8}{x^2}+2x+\dfrac{2}{x}=121+16\\8x^2+\dfrac{8}{x^2}+2x+\dfrac{2}{x}=137$

Taking L.C.M.

$8x^4+8+2x^3+2x=137x^2$

Now,

$8x^4+2x^3-137x^2+2x+8=0$

Here if we substitute the value of $(x=4)$ eqn will become 0.

Hence taking out the factor (x-4) we get:

$(x-4)(4x-1)(2x^2+9x+2)=0$

$2x^2+9x+2=0$

According to the sridharacharya's theorem we get:

$x=\dfrac{-9+\sqrt {65}}{4}\\x=\dfrac{-9-\sqrt {65}}{4}$

And,

$x-4=04x-1=0$

hence:

$x=4\\x=\dfrac{1}{4}$