Let λ=0 be a real number. Let α,β be the roots of 14x2−31x+3λ=0 and α,γ be the roots of 35x2−53x+4λ=0. Then β3α and γ4α are the roots of the equation:
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Official solution
From the first equation: α+β=1431,αβ=143λ.
From the second: α+γ=3553,αγ=354λ.
Dividing the products: γβ=4×143×35=815⟹β=815γ.
Subtracting the sums: β−γ=1431−3553=70155−106=107.
\[\frac{15}8\gamma-\gamma=\frac7{10} \implies \frac78\gamma=\frac7{10} \implies \gamma=\frac45, \qquad \beta=\frac{15}8\cdot\frac45=\frac32.\]
\[\alpha=\frac{31}{14}-\beta=\frac{31}{14}-\frac32=\frac57, \qquad \lambda=\frac{14}3\alpha\beta \implies \lambda=\frac{14}3\cdot\frac57\cdot\frac32=5.\]
Sum of new roots:
\[\frac{3\alpha}\beta+\frac{4\alpha}\gamma=\frac{3\alpha\gamma+4\alpha\beta}{\beta\gamma}=\frac{3\cdot\frac{4\lambda}{35}+4\cdot\frac{3\lambda}{14}}{\frac32\cdot\frac45}=\frac{\frac{12\lambda}{35}+\frac{12\lambda}{14}}{\frac65}=5.\]
Product of new roots:
\[\frac{3\alpha}\beta\cdot\frac{4\alpha}\gamma=\frac{12\alpha^2}{\beta\gamma}=\frac{12\cdot\frac{25}{49}}{\frac65}=\frac{250}{49}.\]
Required equation: x2−5x+49250=0⟹49x2−245x+250=0.