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The roots of the equation $ax^2 +bx+c=0$ where $a,b,c \in\Bbb Z^+$ are $\alpha$ and $\beta$. Find a quadratic with integer coefficients whose roots are $\alpha + \beta$ and $\alpha \beta$.

So my workings are below but from graphing a few curves with my equation, it hasn't seemed to work. Any help would be great.

$$\alpha + \beta = \frac {-b} {a} $$ $$\alpha \beta = \frac c a$$

So the new equation will need to have roots $\frac {-b} {a} $ and $\frac c a$. so the equation can be written in the form $(x-p)(x-q)=0$ meaning the new equation is $$(x+\frac {b} {a})(x-\frac c a) = x^2 +(\frac b a - \frac c a )x - \frac {bc}{a^2}=0$$ $$a^2x+a(b-c)x-bc=0$$ This final answer doesn't seem to yield me the correct result when graphing, any help would be great.

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  • $\begingroup$ Why do you think that the result is wrong? $\endgroup$ – Martin R Feb 22 at 19:13
  • $\begingroup$ When graphing it, it didn't seem to work. Is my logic correct? $\endgroup$ – H.Linkhorn Feb 22 at 19:13
  • $\begingroup$ Of course I may be overlooking something, but I cannot see an obvious error. – Can you provide a concrete example where your result is wrong? $\endgroup$ – Martin R Feb 22 at 19:14
  • $\begingroup$ so if you take $x^2 +3x-10$ which has roots -5 and 2. the equation I've got gives $x^2+13x-30$ which isn't right as it has roots 2 and -15. neither of which are result of summation or multiplication. $\endgroup$ – H.Linkhorn Feb 22 at 19:17
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    $\begingroup$ Ok, maybe I've just made a sign error in my workings somewhere then $\endgroup$ – H.Linkhorn Feb 22 at 19:21
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It is a high-school theorem that, $s$ and $p$ being given, if the sum of two numbers (not necessarily distinct) is $s$ and their product is $p$, these numbers are the roots of the quadratic equation: $$x^2-sx+p=0.$$ So your answer is perfectly correct.

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  • $\begingroup$ I know this, and have used it within my workings but I want an equation which has roots $\alpha + \beta$ and $\alpha \beta$ given the original has roots $\alpha$ and $\beta$ $\endgroup$ – H.Linkhorn Feb 22 at 19:20
  • $\begingroup$ You use this theorem, after you've calculated the sum and the product with Vieta's relations. $\endgroup$ – Bernard Feb 22 at 19:23

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