Question 379391
Use the discriminant.
{{{D=b^2-4ac}}}
If {{{D>0}}}, two real solutions.
If {{{D=0}}}, one real solutions.
If {{{D<0}}}, two complex solutions.
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If the solutions are {{{a}}} and {{{b}}}, then
{{{f(x)=(x-a)(x-b)}}}
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Yes, but they are only different by a constant multiplier. 
{{{f(x)=5(x-2)(x-1)}}}
and 
{{{f(x)=12(x-2)(x-1)}}} both has {{{x=1}}} and {{{x=2}}} but only differ by the constant multiplier in front.
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