Question 1207450
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a,b,c are real numbers
'a' is nonzero


The standard template for a quadratic is
ax^2 + bx + c = 0
Divide everything by 'a' to get
x^2 + (b/a)x + c/a = 0
We'll come back to this later.
The key thing to pay attention to is that the leading coefficient is 1.


Let r and s be the two roots of this quadratic.
If x = r is a root then x-r is a factor
Same goes for the other root.
We find that:
(x-r)(x-s) = 0
which rewrites to
x^2-(r+s)x+rs = 0
Here the leading coefficient is 1 as well.


The product of the roots is the "rs" term at the end.
Compare that to the last term in x^2 + (b/a)x + c/a = 0 to see it matches with the c/a.
Both are constant terms not attached to x in any fashion.


Therefore, rs = c/a


For more information, search out "Vieta's Formulas".


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Alternative route


You can use the quadratic formula like tutor Theo has done, but it could get a bit messy.
The good news is that notice how d = b^2 - 4ac is the discriminant so {{{x = (-b+-sqrt(b^2-4ac))/(2a)}}} would become {{{x = (-b+-sqrt(d))/(2a)}}}


Then that would lead to these roots
{{{x = (-b+sqrt(d))/(2a)}}} and {{{x = (-b-sqrt(d))/(2a)}}}
The numerators multiply to b^2-d when using the difference of squares rule. 
Then compute b^2-d = b^2-(b^2-4ac) = 4ac
The denominators multiply to 4a^2


So this is a fairly quick way to arrive at (4ac)/(4a^2) = c/a.
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