Let , be the roots of the equation x^2-4 x+5=0 and , be the roots of the equation x^2-(3√2…

Question

Let $\alpha,\beta$ be the roots of the equation $x^2-4\lambda x+5=0$ and $\alpha,\gamma$ be the roots of the equation $x^2-(3\sqrt2+2\sqrt3)x+7+3\lambda\sqrt3=0,\ \lambda>0$. If $\beta+\gamma=3\sqrt2$, then $(\alpha+2\beta+\gamma)^2$ is equal to \_\_\_

Answer: 98

Accepted Answer

From the first equation: $\alpha+\beta=4\lambda,\ \alpha\beta=5$. From the second: $\alpha+\gamma=3\sqrt2+2\sqrt3,\ \alpha\gamma=7+3\sqrt3\lambda$.

Since $\alpha$ is common, subtracting the two defining quadratics (evaluated at $\alpha$) and using $\beta+\gamma=3\sqrt2$ leads (after solving the resulting system) to $\lambda=\sqrt2$.

$$\alpha+2\beta+\gamma=(\alpha+\beta)+(\beta+\gamma)=4\lambda+3\sqrt2=4\sqrt2+3\sqrt2=7\sqrt2.$$

$$(\alpha+2\beta+\gamma)^2=(7\sqrt2)^2=98.$$

Solutions · 1