1. A quintic equation!

Solve this quintic equation: $x^5+ax^3 + \frac{1}{5}a^2x + b = 0, \ \ a,b \in \mathbb{R}.$

Note: You don't have to find the roots. Just show us how to find them!

2. Let me try it !

$\cos(5x) = \cos(2x+3x) = \cos(2x)\cos(3x) - \sin(2x)\sin(3x)$

$= ( 2\cos^2(x) - 1)[4\cos^3(x) - 3\cos(x)] - 2\sin(x)\cos(x) \sin(x) [ 3 - 4(1-\cos^2(x))]$

$= = ( 2\cos^2(x) - 1)[4\cos^3(x) - 3\cos(x)] - 2\cos(x)(1- \cos^2(x)) [ 4\cos^2(x) -1 ]$

$= \cos^5(x) [ 8 + 2(4) ] + \cos^3(x)[ -4 -6 -2(4) -2 ] + \cos(x) [ 3 + 2 ]$

$\cos(5x) = 16\cos^5(x) - 20\cos^3(x) + 5\cos(x)$

The next step is substituting $\cos(x) = ku$ , $k$ is a constant

$
16(ku)^5 - 20(ku)^3 + 5(ku) - \cos(5x) = 0
$

$u^5 - \frac{5}{4k^2}u^3 + \frac{5}{16k^4} u - \frac{1}{16k^5}\cos(5x) = 0$

Let $\frac{5}{4k^2} = -a \implies k= \sqrt{\frac{-5}{4a} }$

then let's see what the coefficient of $u$ is

$\frac{5}{ 16 ( \frac{5}{4a} )^2} = \frac{a^2}{5}$

Which equals to that in the equation $u^5 + au^3 + \frac{a^2}{5}u + b = 0$

and finally compare the remaining coefficient : $x^0$

we will see that $- \frac{1}{16k^5}\cos(5x) = b$ so that we can obtain the values of $x$ , the root are $\frac{1}{k}\cos(x_1) , \frac{1}{k}\cos(x_2) , \frac{1}{k}\cos(x_3) ,\frac{1}{k}\cos(x_4) ,\frac{1}{k}\cos(x_5)$

3. Here is another method :

We have $a^5 + b^5 = (a+b)^5 - 5ab(a+b)^3 + 5(ab)^2(a+b)$ or

$(a+b)^5 - 5ab(a+b)^3 + 5(ab)^2(a+b) - (a^5 + b^5) = 0$

Sub. $a+b = x$

$x^5 - 5(ab)x^3 + 5(ab)^2 x - (a^5 +b^5) = 0$

Again , let $ab = \frac{-A}{5}$

we will obtain $x^5 + Ax^3 + \frac{A^2}{5}x - (a^5+b^5) = 0
$

and let $B = -(a^5+b^5)$

$(ab)^5 = - \frac{A^5}{3125}$

Let $\alpha = a^5 , \beta = b^5$

we will find

$\alpha + \beta = -B$ and
$\alpha \beta = - \frac{A^5}{3125}$

we know $\alpha , \beta$ are the roots of the quadratic equation $3125z^2 + 3125B z - A^5 = 0$

After finding them , the roots of the quintic equation can also be found .

For example , let $A = -5 , B= 2$

$\alpha = -1 , \beta = -1$

$a = - 1 , b = -1 \implies x = -2$

which satisfies $(-2)^5 - 5(-2)^3 + 5(-2) + 2 = 0$

4. probably an easier way is to just substitute $x=z-\frac{a}{5z}.$

this problom was solved over 300 years ago by some mathematician!

5. this problom was solved over 300 years ago by some mathematician!
yeah.
Quintic Equation -- from Wolfram MathWorld