Question 84020
This is an EXCELLENT question!  When you solve the quadratic equation {{{ax^2 +bx+c=0}}} if the discriminant {{{b^2-4ac}}} is POSITIVE, then there will be TWO distinct real roots.  If {{{b^2-4ac}}} is NEGATIVE, then there will be NO real roots. If {{{b^2-4ac}}} is ZERO, then there will be only ONE real root, of multiplicity 2.



Now, if you graph the quadratic function {{{y=ax^2 +bx+c}}}, the ROOTS of the quadratic equation described in the previous paragraph correspond to the ZEROS (that is, the x-intercepts) of the graph of the quadratic function.  



Likewise, the NUMBER of roots in the quadratic equation corresponds to the NUMBER of x-intercepts of the quadratic function.  



If the discriminant {{{b^2-4ac}}} of the quadratic function is POSITIVE, then there will be TWO x-intercepts.  For example: {{{y=x^2-4}}}, the discriminant is POSITIVE, since {{{b^2-4ac = 0^2 -4*1*(-4)=16}}}.  There are TWO x-intercepts at x= 2 and at x= -2.
{{{graph(300,300,-10,10,-10,10, x^2-4)}}}. 


If the discriminant {{{b^2-4ac}}} of the quadratic function is NEGATIVE, then there will be NO x-intercepts.  For example: {{{y=x^2+4}}}, the discriminant is NEGATIVE, {{{b^2-4ac = 0^2 -4*1*4=-16}}}.  Notice that there are NO x-intercepts, since the graph does not touch or cross the x-axis.
{{{graph(300,300,-10,10,-10,10, x^2+4)}}}. 



If the discriminant {{{b^2-4ac}}} is ZERO, then there will be only ONE x-intercept, and it will be of multiplicity 2.  For example: {{{y=x^2}}}, the discriminant is ZERO {{{b^2-4ac = 0^2 -4*1*0=0}}}.  There is ONE x-intercept  at x= 0, and it has a multiplicity of 2.
{{{graph(300,300,-10,10,-10,10, x^2)}}}. 


R^2 at SCC