The equation x^5 - 5x^3 + 5x^2 - 1 = 0 has ho | vedclass

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Question

(B) Let $f(x) = x^5 - 5x^3 + 5x^2 - 1$. To find equal roots,we check for common roots between $f(x)$ and its derivative $f'(x) = 5x^4 - 15x^2 + 10x$.

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) Let $f(x) = x^5 - 5x^3 + 5x^2 - 1$. To find equal roots,we check for common roots between $f(x)$ and its derivative $f'(x) = 5x^4 - 15x^2 + 10x$. Setting $f'(x) = 0$: $5x(x^3 - 3x + 2) = 0$. By inspection,$x=1$ is a root of $f'(x)$ since $1-3+2=0$. Checking $f(1) = 1 - 5 + 5 - 1 = 0$. Since $f(1) = 0$ and $f'(1) = 0$,$x=1$ is a repeated root. Dividing $f(x)$ by $(x-1)^2$: $f(x) = (x-1)^2(x^3 + 2x^2 - 2x - 1)$. Checking the derivative of $g(x) = x^3 + 2x^2 - 2x - 1$: $g'(x) = 3x^2 + 4x - 2$. $g(1) = 1 + 2 - 2 - 1 = 0$. Since $g(1) = 0$ and $g'(1) = 3+4-2 = 5 
eq 0$,$x=1$ is a root of multiplicity $3$. Thus,the equation has $3$ equal roots.

The equation $x^5 - 5x^3 + 5x^2 - 1 = 0$ has how many equal roots?

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