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The equation $x^2+ax+b=0$, where $a\neq b$, has solutions $x=a$ and $x=b$. How many such equations are there?

I'm getting $1$ equation as I can only find $a=b=0$ as an equation, which is not allowed.

$$x=\frac{±\sqrt{a^2-4 b}-a}2$$ $x=a$ or $b$ so these are the equations $$a=\frac{\sqrt{a^2-4 b}-a}2$$ $$b=\frac{-\sqrt{a^2-4 b}-a}2$$ $$a=\frac{-\sqrt{a^2-4 b}-a}2$$ $$b=\frac{\sqrt{a^2-4 b}-a}2$$

The only solution for all of these is $a=b=0$, but is this right?

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Hint: Viete's Equations tell us that

$$\begin{cases}ab=b\\{}\\a+b=-a\end{cases}$$

There aren't that many possibilities...(and one of them is with $\;a,b\neq 0\;$)

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You can do this by:

$$(x-a)(x-b)=x^2-(a+b)x+ab=x^2+ax+b$$

Equating constant terms gives $ab=b$ which means $a=1$ or $b=0$

Equating coefficients of $x$ gives $-a-b=a$ or equivalently $2a=-b$

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  • $\begingroup$ I get $\frac{b}{a^2} = -2$, which has infinitely many solutions. $\endgroup$ – SalmonKiller May 12 '15 at 22:19
  • $\begingroup$ @SalmonKiller Not sure how you get that, but note that with $b=ab=a(ab)=a^2b$ you can cancel the $a^2$ and get $b=-2$. You had to know $a\neq 0$ to get that fraction, and $a=b=0$ is an alternative which has to be rejected. $\endgroup$ – Mark Bennet May 13 '15 at 6:01
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Let's take $a=\frac{\sqrt{a^2-4 b}-a}2$ for example:

$$a=\frac{\sqrt{a^2-4 b}-a}2$$ $$2a=\sqrt{a^2-4 b}-a$$ $$3a=\sqrt{a^2-4 b}$$ $$9a^2=a^2-4 b$$ $$8a^2=-4 b$$ $$b=-2 a^2$$

Now you can take that $b$ and substitute in $b=\frac{\sqrt{a^2-4 b}-a}2$ to find a.

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if $x=a$ is a solution to $x^2+ax+b=0$ then $b=-2a^2$

if $x=b$ is a solution $b^2+ab+b=0$ so either $b=0$ or $a+b+1=0$

the second solution gives $=-2a^2+a+1=0$

this equation has solutions $a=1$ and $a=-\frac12$

so it looks like we can have $x^2+x-2$ with solutions of 1 and -2

and $x^2 -\frac12 x -\frac 12$ which has $x=-\frac12$ as a solution.

so $(a,b) = (1,-2)$ and $(a,b) = (-\frac 12,-\frac 12)$ are both valid solutions.

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  • $\begingroup$ (-1/2, -1/2) is a solution, but it is also said in the constraints that $a \neq b$. $\endgroup$ – Dair May 12 '15 at 22:41
  • $\begingroup$ right you are, it seems like the only point of that proviso is to disqualify that solution. $\endgroup$ – WW1 May 12 '15 at 22:45
  • $\begingroup$ Actually I think I see the point, $x^2-\frac12 x -\frac12$ does posses a solution that equals neither $a$ nor $b$, that's why Mark Bennet didn't run across it. $\endgroup$ – WW1 May 12 '15 at 22:56

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