Solving of Equations with Method of Introducing New Variable

It is easier to explain essence of the method of introducing new variable on example.

Example 1. Solve equation `(x^2-3x)^2+3(x^2-3x)-28=0`.

Let `x^2-3x=y`, then initial equation can be rewritten as `y^2+3y-28=0`. This equation has two roots: `y_1=-7,y_2=4`.

Thus, we obtained set of two equations: `x^2-3x=-7,x^2-3x=4`, i.e. `x^2-3x+7=0,x^2-3x-4=0`.

First equation doesn't have roots because its discriminant `D=(-3)^2-4*1*7=-19<0`. Second equation has roots x=4 and x=1. These roots are also roots of initial equation.

Example 2. Solve equation `24/(x^2+2x-8)-15/(x^2+2x-3)=2`.

Let `x^2+2x=y`, then equation can be rewritten as `(24)/(y-8)-(15)/(y-3)=2`.

Common denominator is `(y-8)(y-3)`. So, `(24(y-3))/((y-8)(y-3))-(15(y-8))/((y-3)(y-8))=(2(y-3)(y-8))/((y-3)(y-8))`. This can be rewritten as `(24(y-3)-15(y-8)-2(y-3)(y-8))/((y-8)(y-3))=0`.

Domain of this equation is all y, except y=3 and y=8.

Now, fraction equals zero, when numerator equals zero: `24(y-3)-15(y-8)-2(y-3)(y-8)=0` or `y(2y-31)=0`. Therefore, either `y=0` or `y=31/2`. Both of these roots are in domain of the equation.

But `y=x^2+2x`, so we obtain set of equations: `x^2+2x=0,x^2+2x=31/2`.

First equation has solutions 0 and -2. Second equation has solutions `x=-1+-sqrt(33/2)`.

All these solutions are solutions of initial equation: `0,-2,-1+sqrt(33/2),-1-sqrt(33/2)`.