Question 169705
A quadratic equation will ALWAYS have two solutions.  However, they can be:


2 real and equal solutions (also expressed as one real solution with a multiplicity of 2)


2 real and unequal solutions, or


2 complex solutions


If the discriminant (the part under the radical in the quadratic equation, i.e. the {{{b^2-4ac}}} part) is:


{{{b^2-4ac=0}}}, then you have 2 real and equal solutions


{{{b^2-4ac>0}}},  then you have 2 real and unequal solutions


{{{b^2-4ac<0}}},  then you have a pair of conjugate complex roots of the form:
{{{a+bi}}} and {{{a-bi}}} where {{{i}}} is the imaginary number defined by {{{i^2=-1}}}


If the solutions to a quadratic equation are p and q, then you can write:


{{{(x-p)(x-q)=0}}}, then apply FOIL to multiply the two binomials to obtain your quadratic {{{x^2-(p+q)x+pq=0}}}


There are an infinite number of quadratic equations that can be represented by a given pair of solutions.  {{{x^2-(p+q)x+pq=0}}} is one quadratic with solutions p and q, but {{{2x^2-2(p+q)x+2pq=0}}} is a different quadratic, and {{{ax^2-a(p+q)x+apq=0}}} represents a set of quadratic equations with a set element, each different from the others, for every possible value of a.


Example (real/unequal):


Let {{{x[1]=2}}} and {{{x[2]=-1}}} be solutions of a particular quadratic equation.


Then {{{(x-2)(x+1)=x^2-x-2=0}}} is a quadratic equation with the given solutions as roots.


Example (real/equal):


Let {{{x[1]=2}}} and {{{x[2]=2}}} be solutions of a particular quadratic equation.


Then {{{(x-2)(x-2)=x^2-4x+4=0}}} is a quadratic equation with the given solutions as roots.


Example (complex)


Let {{{x[1]=1+i}}} and {{{x[2]=1-i}}} be solutions of a particular quadratic equation.


Then {{{(x-(1+i))(x-(1-i))=(x-1-i)(x-1+i)=x^2-x+ix-x+1-i-ix+i+1=x^2-2x+2=0}}} is a quadratic equation with the given solutions as roots.


Hope that helps