Solving equation reducible to quadratic form

Examples

Here's an example of solving a quadratic equation of the form $ap(x)+\frac b{p(x)}=c$ where $a,b$ and $c$ are real numbers, $a≠0$ and $p(x)$ is an algebraic expression .

Solve the equation: $2(\frac x{x+1})^2-5(\frac x{x+1})+2 = x\neq -1$

Explanation:

Given equation: $2(\frac x{x+1})^2-5(\frac x{x+1})+2 = x\neq -1$

Step 1: Let $(\frac x{x+1})=y$

to get a quadratic equation in terms of $y$: $2(y)^2 - 5(y) +2$
or

$2y^2 - 5y +2 =0$

Step 4: Factor the quadratic equation or quadratic formula can be applied:

$2y^2 - 5y +2 =2y^2 -4y-y+2=0$ $2y(y-2)-1(y-2)=0\\(2y-1)(y-2)=0$

Step 5: Solve for $y$ by setting each factor equal to zero: $(2y-1)(y-2)=0 \\2y-1=0 \space or \space y-2=0$ $y=2$ or $y=\frac12$

Step 6: Substitute back $\frac x{x+1}$ for $y$ and solve for $x$:

$\frac x{x+1}=y$ $\frac x{x+1}=2$ or $\frac x{x+1}=\frac12$

$x=2(x+1)$ or $2x=x+1$

$x=2x+2$ or $2x-x=1$

$x=-2$ or $x=1$

Solution:

Given equation: $2(\frac x{x+1})^2-5(\frac x{x+1})+2 = x\neq -1$

Let $(\frac x{x+1})=y$

so given equation will become $2(y)^2 - 5(y) +2$
or

$2y^2 - 5y +2 =0$

$2y^2 - 5y +2 =2y^2 -4y-y+2=0$ $2y(y-2)-1(y-2)=0\\(2y-1)(y-2)=0$

$\\2y-1=0 \space or \space y-2=0$ $y=2$ or $y=\frac12$

Since, $\frac x{x+1}=y$ substituting back: $\frac x{x+1}=2$ or $\frac x{x+1}=\frac12$

$x=2(x+1)$ or $2x=x+1$

$x=2x+2$ or $2x-x=1$

$x=-2$ or $x=1$

****The solution set is $\{-2,1\}$