Finding the value of $p$ given the roots of a quadratic equation

Question

Could you tell me how could I solve the value of $p$ in this question?

For which $p$ does $3x^2+(p+1)x+24=0$ have one root equal to twice the other root? Options given are $\{\pm17,\pm19\}$ with all possible sign combinations.

Answer 1

On one hand, $$3x^2+(p+1)x+24=0$$ $$x^2+\frac{p+1}3x+8=0$$ On the other hand, $$(x-r)(x-2r)=x^2-3rx+2r^2=0$$ Comparing constant coefficients we see that $r=\pm2$, from which (comparing linear coefficients) we get $\frac{p+1}3=\pm6$. This yields $p=+17$ and $p=-19$.