Question 173956
I don't think you stated your initial question precisely.  I think what you meant to ask is "How do you know if a quadratic equation will have one, two, or no <i><b>real number</b></i> solutions?


In fact, the fundamental theorem of algebra tells us that a polynomial of degree n has n roots.  Therefore a quadratic, which is a polynomial of degree 2, must ALWAYS have two roots.  However, sometimes the roots are complex numbers so when you restrict your definition of 'solution' to the real numbers, you can have a 'no solution' result.


As to the question about one root, that is still a subject of some debate.  Some say that a quadratic always has two roots, but in the case of a perfect square trinomial, those two roots will be identical.  Others say that there is a single root with a multiplicity of two for that situation.


The process to determine the character of the roots of a quadratic is as follows:

Step 1:  Put the quadratic into standard form, namely {{{ax^2+bx+c=0}}}


Step 2:  Calculate the discriminant {{{DELTA}}}, which is the expression under the radical in the quadratic formula {{{x = (-b +- sqrt( b^2-4ac ))/(2a) }}}, namely:  {{{DELTA=b^2-4ac}}}.


If {{{DELTA = 0}}} then you either have two real and identical roots, or you have one root with a multiplicity of two, depending on your school of thought.  This is your 'one solution' situation.


If {{{DELTA > 0}}}, then you have two different real number roots.  This is your 'two solutions' situation.


If {{{DELTA < 0}}}, then you have a conjugate pair of complex number roots of the form {{{a+-bi}}} where {{{i}}} is the imaginary number defined by {{{i^2=-1}}}.  This is your 'no solution' situation.


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Given two numbers {{{p}}} and {{{q}}}, you can derive a quadratic polynomial with those two numbers as roots by using the fact that {{{a}}} is a root of a polynomial in {{{x}}} if and only if {{{x-a}}} is a factor of the polynomial.


Therefore, given {{{p}}} and {{{q}}} as the two roots of a quadratic, one of the quadratic polynomials with {{{p}}} and {{{q}}} as roots can be derived from {{{(x-p)(x-q)=x^2-(p+q)x+pq}}}.


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Your last question, "Is it possible to have different quadratic equations with the same solution?"


You need to be very careful with your language here.


{{{x^2-x-6=0}}} has a solution set {-2,3}, but so does {{{-x^2+x+6=0}}}.  However, one could successfully argue that these two equations are equivalent, and therefore are not really different quadratic equations.  So, the answer to your question as stated is no, you can't have different quadradic EQUATIONS with the same solution.


But do not confuse a quadratic equation {{{ax^2+bx+c=0}}} with a polynomial function of degree 2:  {{{f(x)=ax^2+bx+c}}}.  Looking at the previous example, {{{f(x)=x^2-x-6}}} and {{{f(x)=-x^2+x+6}}} are two very different functions -- one is a parabola opening upwards and the other is a parabola opening downwards.