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

 Jun 11, 2024
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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

 Jun 11, 2024

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