Note: For a,b,c∈R and a≠0, if b=0 in the equation ax²+bx+c=0, the solution set is found by in the following way: ax²+c=0, after we get ax²=-c, then divide both side with a; x²=-c/a
If -c/a>0, the roots of the given equation are
x1= √-c/a and x2=√-c/a
If -c/a<0, there is no real roots of the given equation on R. It is stated SS=Ø
