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

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.

• There are two possible answers I assumed x1=y and x2=2y – Autumn Fall Sep 18 at 4:43
• and what did you get? You should post your efforts in your question body. "Here's my homework, do it for me" questions do very poorly on this site. – YiFan Sep 18 at 4:45

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 $$p=-19$$.