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# Algebra

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

Aug 16, 2024

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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! :)

Aug 16, 2024
edited by NotThatSmart  Aug 16, 2024
edited by NotThatSmart  Aug 16, 2024