# Prove, by contradiction, that a quadratic equation, $ax^2 + bx +$c where $a\neq 0$, has at most two distinct roots.

My approach was to show that the discriminant is less than zero, i.e, there are indeed no real roots, which contradicts that there are at most two distinct roots. I'm not exactly sure how to execute this theory however. Should I do so by introducing constants which would make the discriminant less than zero? Proofs is not my strong point :(

Help greatly appreciated.

• For contradiction, you would assume you have three or more roots. Now maybe you can get a contradiction from either the quadratic formula or some factorization theorem, whatever you have at your disposal. – Randall Nov 3 '17 at 14:35
• @Randall I don't see how that is possible as you can only show there are at most two distinct roots either way. – blazedtrailz Nov 3 '17 at 14:42
• Right, it is simpler via a direct proof. You've asked for one by contradiction, though. – Randall Nov 3 '17 at 14:44
• I agree, proving this by contradiction seems strange since you can just calculate the roots. – RoyPJ Nov 3 '17 at 14:50
• "there are indeed no real roots, which contradicts that there are at most two distinct roots": hem, given that $0\le2$, I don't see a contradiction. – Yves Daoust Nov 3 '17 at 15:38

First of all solve the equation for its two roots, hence

$x_1 = \frac{-b -\sqrt{b^2-4ac}}{2a}; x_2 = \frac{-b +\sqrt{b^2-4ac}}{2a}$. (Assume those roots exist in real numbers, otherwise there are less then 2 anyway).

Now you can write your equation as $ax^2 +bx+c = a(x-x_1)(x-x_2)$.

Assume there is a $x_3 \neq x_2,x_1$ with $ax_3^2+bx_3+c=0$.

A real product is zero if and only if (at least) one of its factors is zero. Therefore

$a(x-x_1)(x-x_2) = 0 \Leftrightarrow (x=x_1 \vee x=x_2)$

It follows that $ax_3^2+bx_3+c \neq 0$ what contradicts to the assumption.

• Of course, sorry for the mistake, I corrected it – RoyPJ Nov 3 '17 at 15:04
• Not a valid argument. Knowing two roots allows you to write $ax^2+bx+c=p(x)(x-x_1)(x-x_2)$, but $p(x)$ can still have roots. You are essentially saying "I know the two roots, they are this and this, so there can't be a third one". – Yves Daoust Nov 3 '17 at 15:24
• I don't see your point. You just divide by $a$, then solve it, then you can conclude $x^2+\frac{b}{a}x + \frac{c}{a} = (x-x_1)(x-x_2)$ and multiply both sides with $a$? – RoyPJ Nov 3 '17 at 15:33
• Yep, "solve it" exactly means that you know there are two roots. If you allow yourself to do this, just invoke the fundamental theorem of algebra and you are done. – Yves Daoust Nov 3 '17 at 15:36
• First I assumed two roots exist. Okay, then replace "solve it" through "guess two roots and prove them". Pretty sure this doesn't break the fundamental theorem of Algebra. Your argument is completely invalid anyway since if you can compute two roots, then they exist. – RoyPJ Nov 3 '17 at 15:43

You cannot conclude that there are no real roots, because it is possible (for example, if $a=1,c=-1$) that there are real roots.

The most basic way of proving your problem is to "complete the square". Let me start for you. You can write $$ax^2+bx+c=\left(\sqrt{a}x+\frac{b}{2}\right)^2 - D$$ for some value $D$ (which you should try to calculate).

Then, if $D<0$, clearly, the expression has no roots (because it is something positive plus something squared, so it is positive). If $D\geq 0$, then you can use the rule $$a^2-b^2=(a-b)(a+b)$$ to simplify the expression.

Once you have the expression in the form of $$A(x-a_1)(x-a_2)$$

it should be easy to conclude that it can equal zero only if $x\in\{a_1,a_2\}$

• Where is the contradiction ? – Yves Daoust Nov 3 '17 at 15:25

Assume that there is at least one real root $x_1$. Then $$f(x)=a^2+bx+c-(ax_1^2+bx_1+c)=(x-x_1)(a(x+x_1)+b)$$ As the factorization is complete, you can only have roots at the roots of the linear factors, other than $x_1$ that is $-\frac{b}a-x_1$.