2020 AIME II Problems/Problem 11

Solution 3

We know that $P(x)=x^2-3x-7$ .

Since $Q(0)=2$ , the constant term in $Q(x)$ is $2$ . Let $Q(x)=x^2+ax+2$ .

Finally, let $R(x)=x^2+bx+c$ .

$P(x)+Q(x)=2x^2+(a-3)x-5$ . Let its roots be $p$ and $q$ .

$P(x)+R(x)=2x^2+(b-3)x+(c-7)$ Let its roots be $p$ and $r$ .

$Q(x)+R(x)=2x^2+(a+b)x+(c+2)$ . Let its roots be $q$ and $r$ .

By vietas, $p+q=\frac{3-a}{2}, p+r=\frac{3-b}{2}, q+r=\frac{-(a+b)}{2}$

We could work out the system of equations, but it's pretty easy to see that $p=\frac32, q=-\frac{a}{2}, r=-\frac{b}{2}$ .

$\text{Again, by vietas, }pq=-\frac52\text{, } pr=\frac{c-7}{2}\text{, } qr=\frac{c+2}{2}$ $\text{Multiplying everything together a}\text{nd then taking the sqrt of both sides,}$ $(pqr)^2=\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)$ $pqr=\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)}$ $\text{Now, we divide this }\text{equation by }qr=\frac{c+2}{2}$ $\frac{pqr}{qr}=\frac{\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)\left(\frac{c+2}{2}\right)}}{\frac{c+2}{2}}$ $p = \frac{\sqrt{\left(-\frac52\right)\left(\frac{c-7}{2}\right)}}{\sqrt{\frac{c+2}{2}}}$ $\text{Recall th}\text{at }p=\frac32 \text{ and square both sides}$ $\frac94=\frac{\left(-\frac52\right)\left(\frac{c-7}{2}\right)}{\frac{c+2}{2}}$ $\text{Solving gives } c=\frac{52}{19}, \text{ so our answer is }\boxed{071}$ \end{align*}

~quacker88

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2020 AIME II Problems/Problem 11

Solution 3