Q.Solve the equation $x^3-9x^2+14x+24=0$ if it is given that two of its roots are in the ratio $3:2$.

Step 1. Set up. Let two roots be $3k,2k$ (ratio $3:2$) and the third be $\gamma$. For $x^3-9x^2+14x+24=0$: sum $=9$, pairwise sum $=14$, product $=-24$.

Step 2. Sum relation. $3k+2k+\gamma=9 \implies \gamma=9-5k$.

Step 3. Pairwise-sum relation. $3k(2k)+2k\gamma+\gamma(3k)=14 \implies 6k^2+5k\gamma=14$. Substituting $\gamma=9-5k$: $6k^2+5k(9-5k)=14 \implies 6k^2+45k-25k^2=14 \implies -19k^2+45k-14=0 \implies 19k^2-45k+14=0$.

Step 4. Solve for $k$. $\Delta=45^2-4(19)(14)=2025-1064=961=31^2$. $k=\dfrac{45\pm31}{38}$, giving $k=2$ or $k=\tfrac{7}{19}$; take the value giving integer roots, $k=2$. …

Assume the ratio-constrained roots as $3k,2k$, combine with Vieta' …

• Forgetting to check the found roots satisfy ALL THREE Vieta relations, not just the two used to so …