+2653 Find the value of $v$ such that $\frac{-21-\sqrt{201}}{10}$ a root of $5x^2+21x+v = 0$.
+135 We can use the Quadratic formula to find the value of v.
a = 5, b = 21, c = v
\(\frac{-21-\sqrt{201}}{10} = {-b \pm \sqrt{b^2-4ac} \over 2a}\)
\(\frac{-21-\sqrt{201}}{10} = {-21 \pm \sqrt{21^2-4(5)(v)} \over 2(5)}\)
\(\frac{-21-\sqrt{201}}{10} = {-21 - \sqrt{441-20v} \over 10}\)
\(\sqrt{201} = \sqrt{441-20v}\)
\(201 = 441-20v\)
\(20v = 240\)
\(\mathbf{v=12}\)
Therefore, the value of v is 12
