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.
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.