Parametric Equations

Suppose we are given equality with variable x and a: `f(x,a)=0`. If we need for real value of a solve this equation with respect to x, then equation `f(x,a)=0` is called equation with variable x and parameter a. To solve equation with parameter a means for every a to find x, that satisfy given equation.

Example 1. Solve equation `2a(a-2)x=a-2`.

First of all we need to consider those values of parameter that make coeffcient near x equal zero.

So, `2a(a-2)=0` when a=0 or a=2.

If a=0 then equation becomes `2*0*(0-2)x=0-2` or `0=-2`. This equation dosn't have roots.

If a=2 then equation becomes `2*2*(2-2)x=2-2` or `0=0`. Any real number is root of this equation, i.e. there are infinitely many roots.

Now, consider case when `2a(a-2)!=0`: in this case we can divide both sides of equation by `a(a-2)` and obtain following: `x=(a-2)/(2a(a-2))=1/(2a)`.

So, answer is follwoing: if a=0, then there are no roots, if a=2 then any real number of the equation, if `a!=0` and `a!=2` then there is one root `x=1/(2a)`.

Example 2. Solve equation `(a-1)x^2+2(2a+1)x+4a+3=0`.

If a=1, then `a-1=0` and equation is not qudratic, it is linear. So if a=1 then `(1-1)x^2+2(2*1+1)x+4*1+3=0` or `6x+7=0`. From this we have that `x=-7/6`.

If `a!=1` then `a-1!=0` and we have quadratic equation. Its discriminant is `D=(2(2a+1))^2-4((a-1)(4a+3))=4(4a^2+4a+1)-4(4a^2-a-3)=`

`=4(5a+4)`.

If `D<0`, i.e. `5a+4<0` or `a<-4/5` then equation doesn't have roots.

If `D=0`, i.e. `5a+4=0` or `a=-4/5` then equation has one root `x=(-2(2a+1))/(2*(a-1))=-(2a+1)/(a-1)=-(2*(-4/5)+1)/(-4/5-1)=-1/3`.

If `D>0`, i.e. `5a+4>0` or `a> -4/5` and `a!=1` then equation has two roots `x=(-2(2a+1)+-sqrt(4(5a+4)))/(2*(a-1))=(-(2a+1)+-sqrt(5a+4))/(a-1)`.

So, answer is following:

  1. If `a<-4/5` then there are no roots;
  2. If `a=-4/5` then `x=-1/3`;
  3. If `a=1` then `x=-7/6`;
  4. If `a> -4/5` and `a!=1` then `x=(-(2a+1)+-sqrt(5a+4))/(a-1)`.