2. (15 points). Form a quadratic equation whose roots are twice the roots of the equation $2 x^{2}-5 x-8=0$.

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Solution: According to Vieta's theorem, $x_{1}+x_{2}=\frac{5}{2}, x_{1} \cdot x_{2}=-4$. Let $t_{1}$ and $t_{2}$ be the roots of the desired quadratic equation, then $t_{1}+t_{2}=2 x_{1}+2 x_{2}=5, t_{1} \cdot t_{2}=2 x_{1} \cdot 2 x_{2}=-16$ and by the inverse Vieta's theorem we get the equation: $x^{2}-5 x-16=0$. The equation is a monic polynomial.

Answer: $x^{2}-5 x-16=0$.

Criteria:

| Points | Grading Criteria |
| :---: | :--- |
| 10 | Correct answer obtained with justification |
| 5 | Incorrect answer due to an arithmetic error |
| 0 | The solution does not meet any of the above conditions |