Solve-x-4-x-3-2ax-2-ax-a-2-0-a-R-

Question Number 70399 by Henri Boucatchou last updated on 04/Oct/19

S o l v e x^{4} + x^{3} - 2 {a x}^{2} - a x + a^{2} = 0, a \in R

Commented by Prithwish sen last updated on 04/Oct/19

x^{4} - {2 ax}^{2} + a^{2} + x^{3} - ax = 0

{(x^{2} - a)}^{2} + x (x^{2} - a) = 0

(x^{2} - a) (x^{2} + x - a) = 0

\Rightarrow x = \pm \sqrt{} a and x = \frac{- 1 \pm \sqrt{1 + 4 a}}{2}

I have not checked the answer . Please check it .

Commented by Prithwish sen last updated on 04/Oct/19

Thank you very much Sir . I correct it .

Commented by MJS last updated on 04/Oct/19

you forgot the “ {\sqrt{}}^{″}

Answered by MJS last updated on 04/Oct/19

trying factors of a^{2}

x \in \pm {a^{2}, a, \sqrt{a}}

\Rightarrow x = \pm \sqrt{a}

\Rightarrow

x^{4} + x^{3} - 2 {a x}^{2} - a x + a^{2} =

= (x - \sqrt{a}) (x + \sqrt{a}) (x^{2} + x - a) =

= (x - \sqrt{a}) (x + \sqrt{a}) (x + \frac{1}{2} - \frac{\sqrt{4 a + 1}}{2}) (x + \frac{1}{2} + \frac{\sqrt{4 a + 1}}{2})

\Rightarrow x = \pm \sqrt{a} \lor x = - \frac{1}{2} \pm \frac{\sqrt{4 a + 1}}{2}

\Rightarrow {\begin{cases} 4 real solutions for 0 ⩽ a \\ 2 real and 2 imaginary solutions for - \frac{1}{4} ⩽ a < 0 \\ 2 conjugated complex and 2 imaginary solutions for a < - \frac{1}{4} \end{cases}

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