Find the value of $v$ such that $\frac{-21-\sqrt{201}}{10}$ a root of $5x^2+21x+v = 0$.

Answer 1

We don't actually have to evaluate the expression $\frac{-21 - \sqrt{201}}{10}$ to solve the problem. Since it is a root of the quadratic equation $5x^2 + 21x + v = 0$, we must have that $\frac{-21 - \sqrt{201}}{10}$ is equal to one of the values of the roots of the equation. But the discriminant of the quadratic equation $5x^2 + 21x + v = 0$ is $21^2 - 4 \cdot 5 \cdot v = 441 - 20v$, and the quadratic formula tells us that the roots are \begin{align*}
x=\frac{-21 \pm \sqrt{441 - 20v}}{10}.
\end{align*} Thus, we want this expression to equal $\frac{-21 - \sqrt{201}}{10}$. So we set \begin{align*}
\frac{-21 + \sqrt{441 - 20v}}{10} &= \frac{-21 - \sqrt{201}}{10} \\
-21 + \sqrt{441 - 20v} &= -21 - \sqrt{201} \\
2\sqrt{441 - 20v} &= \sqrt{201} \\
4(441 - 20v) &= (\sqrt{201})^2 \\
4(441 - 20v) &= 201 \\
1764 - 80v &= 201 \\
80v &= 1563 \\
v &= \boxed{\frac{1563}{80}}.
\end{align*}