Algebra

We can complete this problem in two different ways. 

The first tactic is to essentially compare this root to the quadratic equation of \(5x^2+21x+v = 0\)

From this quadratic, we can identify that a = 5, b = 21, c = v. 

We have the equation

\(\frac{-21-\sqrt{201}}{10} = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ \frac{-21-\sqrt{201}}{10} = {-21 \pm \sqrt{21^2-4(5)(v)} \over 2(5)}\\ \frac{-21-\sqrt{201}}{10} = {-21 - \sqrt{441-20v} \over 10}\\ \sqrt{201} = \sqrt{441-20v}\\ 201 = 441-20v\\ v=12\)

This is a bit complicated and takes a lot of computations, but it does give us the correct answer. 

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The second tactic is to use conjugations of square roots. 

This is because the conjugate root theorem states that if a root of a polynomial is a square root \(a+\sqrt b\), then its conjugate, \(a-\sqrt b\) is also a root

We can apply that to this problem. If  \(\frac{-21-\sqrt{201}}{10}\) is a root, then \(\frac{-21 + \sqrt {201}}{ 10 }\) is also a root. 

The product of the roots is \([ (-21)^2 - 201 ] / 100 = 240/100 = 2.4\)

However, in the quadratic, we also have that \(v / 5\)

Thus, we have 

\(v/5 = 2.4\\ v = 12\)

SO 12 is the final answer. 

Thanks! :)