# Thread: [SOLVED] How to solve this equation?

1. ## [SOLVED] How to solve this equation?

By using the substitution $\displaystyle u = \frac{1}{x} + x$, show that the equation $\displaystyle 6x^4 - 25x^3 + 12x^2 - 25x + 6 = 0$ can be expressed as $\displaystyle 6u^2 - 25u = 0$. Hence, solve the equation giving your answer correct to two decimal places.

2. $\displaystyle 6u^2-25u=0$ --------- (1)

Substituting $\displaystyle u=x+\frac{1}{x}$ in (1),

$\displaystyle 6(x+\frac{1}{x})^2-25(x+\frac{1}{x})=0$

$\displaystyle 6(x^2+2+\frac{1}{x^2})-25(x+\frac{1}{x})=0$

$\displaystyle 6(x^4+2x^2+1)-25(x^3+x)=0$

$\displaystyle 6x^4-25x^3+12x^2-25x+6=0$

3. How about the roots? How to solve the roots? The answers given in the book are 3.91, 0.26, i and -i. How did they got it? Showing steps are very much welcomed. Thanks.

4. $\displaystyle 6u^2-25u=0$

$\displaystyle u(6u-25)=0$

$\displaystyle u=0$ or $\displaystyle u=\frac{25}{6}$

1. $\displaystyle u=0$

$\displaystyle x+\frac{1}{x}=0$

$\displaystyle x=-\frac{1}{x}$

$\displaystyle x^2=-1$

$\displaystyle x=\pm i$

2. $\displaystyle u=\frac{25}{6}$

$\displaystyle x+\frac{1}{x}=\frac{25}{6}$

$\displaystyle \frac{x^2+1}{x}=\frac{25}{6}$

$\displaystyle 6x^2-25x+6=0$

$\displaystyle x=\frac{25\pm \sqrt{625-144}}{12}$

$\displaystyle x=\frac{25\pm \sqrt{481}}{12}$

$\displaystyle x=\frac{25\pm 21.93}{12}$

$\displaystyle x=3.91$ or $\displaystyle 0.26$

5. Hello, mark1950!

This requires Olympic-level gymnastics . . .

By using the substitution $\displaystyle u = x + \frac{1}{x}$ .show that the equation:

$\displaystyle 6x^4 - 25x^3 + 12x^2 - 25x + 6 \:=\: 0$ can be expressed as $\displaystyle 6u^2 - 25u \:=\: 0$.

Hence, solve the equation, giving your answer correct to two decimal places.

We have: .$\displaystyle (6x^4 + 12x^2 + 6) - (25x^3 + 25x) \:=\:0 \quad\Rightarrow\quad 6(x^4 + 2x + 1) - 25(x^3+x) \:=\:0$

Divide by $\displaystyle x^2\!:\;\;6\left(x^2 + 2 + \frac{1}{x^2}\right) - 25\left(x + \frac{1}{x}\right) \:=\:0 \quad\Rightarrow\quad 6\left(x + \frac{1}{x}\right)^2 - 25\left(x + \frac{1}{x}\right) \:=\:0$

Let .$\displaystyle u \:=\:x+\frac{1}{x}\!:\quad 6u^2 - 25u \:=\:0 \quad\Rightarrow\quad u(6u-25) \:=\:0$

. . And we have: .$\displaystyle u \:=\:0, \;u \:=\:\frac{25}{6}$

Back-substitute:

$\displaystyle u \:=\:0 \quad\Rightarrow\quad x + \frac{1}{x} \:=\:0 \quad\Rightarrow\quad x^2 + 1 \:=\:0\quad\Rightarrow\quad x^2 \:=\:-1$ . . . no real roots

$\displaystyle u \:=\:\frac{25}{6} \quad\Rightarrow\quad x + \frac{1}{x} \:=\:\frac{25}{6} \quad\Rightarrow\quad 6x^2 + 6 \:=\:25x \quad\Rightarrow\quad 6x^2 - 25x + 6 \:=\:0$

. . Quadratic Formula: .$\displaystyle x \;=\;\frac{25 \pm\sqrt{481}}{12} \;\approx\;\begin{Bmatrix}3.91 \\ 0.26\end{Bmatrix}$

6. Thanks a lot, guys! You really helped!