Question 154861
How many solutions exist for a quadratic equation?  
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The number of roots of a polynomial equation is equal to the degree of the polynomial (the exponent of the leading term). 
Quadratic equations are of degree 2, {{{x^2}}}.
They have two (2) roots.
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How do we detemine whether the solutions are real or complex? 
Use the discriminant.
For the general quadratic equation, 
{{{ax^2+bx+c=0}}}
the discriminant is 
{{{D=b^2-4ac}}}
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If {{{D>0}}} then you have two distinct real roots.
Example:
{{{x^2-7x+10=(x-2)(x-5)}}}
{{{D=49-4(10)=9}}}
2 real roots, x=2,5 .
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If {{{D=0}}}, you have a double root, one real root occurring twice 
{{{x^2-2x+1=(x-1)^2}}}
{{{D=4-4(1)=0}}}
2 real roots, x=1,1.
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If {{{D<0}}}, you have two complex roots, that are complex conjugates.
{{{x^2+1=(x+i)(x-i)}}}
{{{D=0-4(1)=-4}}}
2 complex roots, x=i,-i.