# 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).

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).