MHB How Does a Double Root Lead to the Equation $108a^5=-3125b^3$?

[anemone]

Here is this week's POTW:

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The equation $x^5=ax^2+b$ where $b\ne 0$ has a double root. Show that $108a^5=-3125b^3$.

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[anemone]

Congratulations to the following members for their correct solution:):

1. kaliprasad
2. greg1313
2. lfdahl

Solution from greg1313:

Spoiler

Let the double root be $$q$$. Then we have

$$(x^2-2qx+q^2)(x^3+Bx^2+Cx+D)$$

$$=x^5-2qx^4+q^2x^3+Bx^4-2Bqx^3+Bq^2x^2+Cx^3-2Cqx^2+Cq^2x+Dx^2-2Dqx+Dq^2$$

Equating coefficients we have

$$B=2q,\quad C=3q^2,\,\quad D=\dfrac32q^3$$

thus

$$a=Bq^2-2Cq+D=-\dfrac52q^3,\quad b=Dq^2=\dfrac32q^5$$

and

$$108\left(-\dfrac52q^3\right)^5=\dfrac{27}{8}\cdot-3125q^{15}$$

$$-3125\left(\dfrac32q^5\right)^3=\dfrac{27}{8}\cdot-3125q^{15}$$

as required.

Alternate solution from lfdahl:

Spoiler

Let $r$ denote the double root in $x^5=ax^2+b$ , $b \ne 0 \;\;\;\; $ (1).

Then $r$ is also root in the first derivative of (1), which gives us two equations to follow:

$r^5=ar^2+b \Rightarrow r^2(r^3-a) = b \Rightarrow r^6(r^3-a)^3 = b^3 \;\;\;\;$ (2).

and

$5r^4 = 2ar \Rightarrow r^3 = \frac{2}{5}a \;\;\;\; $ ($r \ne 0$, because $b \ne 0$) $\;\;\;\;$ (3).Equation (3) inserted in (2) yields:$(\frac{2}{5}a)^2(-\frac{3}{5}a)^3=b^3 \Rightarrow 3^3\cdot 2^2 \cdot a^5 = -5^5 \cdot b^3 \Rightarrow 108a^5 = -3125b^3$.