Polynomials Question (1 Viewer)

Ambility · Feb 27, 2016

How would you do this?

$$The polynomial equation $x^5-tx^2+q=0$ where $t$ and $q$ are constants, has a multiple root. Show that $108t^5=3125q^3$

Paradoxica · Feb 27, 2016

Ambility said:
How would you do this?

$$\noindent The polynomial equation $x^5-tx^2+q=0$ where $t$ and $q$ are constants, has a multiple root. Show that $108t^5=3125q^3$

$$\noindent Differentiating the polynomial yields $5x^4 -2tx = 0 \\ $It is clear $0$ is not a root of the original polynomial.$\\ 5x^3 = 2t \Leftrightarrow x^2 = \left(\frac{2t}{5}\right)^{\frac{2}{3}} \\ \left( -\frac{3t}{5}\right) \left(\frac{2t}{5}\right)^{\frac{2}{3}} +q =0\\\left(\frac{3t}{5}\right) \left(\frac{2t}{5}\right)^{\frac{2}{3}}=q \\ \left(\frac{27t^3}{125}\right)\left(\frac{4t^2}{25}\right)=q^3 \\ 108t^5 = 3125q^3$

Ambility · Feb 28, 2016

Paradoxica said:
$$\noindent Differentiating the polynomial yields $5x^4 -2tx = 0 \\ $It is clear $0$ is not a root of the original polynomial.$\\ 5x^3 = 2t \Leftrightarrow x^2 = \left(\frac{2t}{5}\right)^{\frac{2}{3}} \\ \left( -\frac{3t}{5}\right) \left(\frac{2t}{5}\right)^{\frac{2}{3}} +q =0\\\left(\frac{3t}{5}\right) \left(\frac{2t}{5}\right)^{\frac{2}{3}}=q \\ \left(\frac{27t^3}{125}\right)\left(\frac{4t^2}{25}\right)=q^3 \\ 108t^5 = 3125q^3$

Where did you get

$\left( -\frac{3t}{5}\right) \left(\frac{2t}{5}\right)^{\frac{2}{3}} +q =0$

from?

Paradoxica · Feb 28, 2016

Ambility said:
Where did you get

$\left( -\frac{3t}{5}\right) \left(\frac{2t}{5}\right)^{\frac{2}{3}} +q =0$

from?

Solve for the only real, non-zero, root of the derivative. That is the root of the original equation. (presumably, not sure what happens of you pick the complex roots.)

Ambility · Feb 28, 2016

Paradoxica said:
Solve for the only real, non-zero, root of the derivative. That is the root of the original equation. (presumably, not sure what happens of you pick the complex roots.)

Right, I get how you found the root of the derivative, or the square of the root for that matter. Did you get the quoted line from subbing that back into the original equation? If so, shouldn't there be 3 terms seeing as $x^5-tx^2+q=0$ is the original equation?

Paradoxica · Feb 28, 2016

Ambility said:
Right, I get how you found the root of the derivative, or the square of the root for that matter. Did you get the quoted line from subbing that back into the original equation? If so, shouldn't there be 3 terms seeing as $x^5-tx^2+q=0$ is the original equation?

Not if you simplify.

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