Suppose $a$ is a real number for which all the roots of the equation $x^{4} - 2ax^{2}+x+a^{2}-a=0$ are real. Then
(a) $\quad a<-\frac{2}{3}$
(b) $\quad a=0$
(c) $\quad 0<a<\frac{3}{4}$
(d) $\quad a\ge \frac{3}{4}$
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2For these problems that are multiple choice, it suffices to simply choose a few values. For example, it's easy to see this won't be true if $a=0$. – Simply Beautiful Art Mar 01 '17 at 14:14
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I guess that's true, since all of the options have am empty intersection, it should be fine. – Naweed G. Seldon Mar 01 '17 at 14:30
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Of course, choose easy values, like $a=1,-1$... – Simply Beautiful Art Mar 01 '17 at 14:31
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This question can be solved without having to do much algebra if you know what the graph of the quartic $f(x) = (x^2-a)^2$ looks like. Maybe I'll write an answer up later (drawing pictures on computer takes some time). – Josh Mar 01 '17 at 14:56
1 Answers
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Solve $a$ in terms of $x$ first,
\begin{align*} a^2-(2x^2+1)a+x^4+x &= 0 \\ a &= \frac{2x^2+1\pm \sqrt{(2x^2+1)^2-4(x^4+x)}}{2} \\ &= \frac{2x^2+1 \pm \sqrt{4x^2-4x+1}}{2} \\ &= \frac{2x^2+1 \pm (2x-1)}{2} \\ &= x^2 \pm x+\frac{1\mp 1}{2} \\ &= x^2+x \qquad \text{or} \qquad x^2-x+1 \\ x^4-2ax^2+x+a^2-a &= (x^2+x-a)(x^2-x-a+1) \end{align*}
Case I: $x^2+x-a=0$ \begin{align*} \Delta &\ge 0 \\ 1+4a &\ge 0 \\ a &\ge -\frac{1}{4} \end{align*}
Case II: $x^2-x-a+1=0$ \begin{align*} \Delta &\ge 0 \\ 1+4(a-1) &\ge 0 \\ a &\ge \frac{3}{4} \end{align*}
For all roots to be real,
$$a\in \left[ -\frac{1}{4}, \infty \right) \cap \left[ \frac{3}{4}, \infty \right)$$
$$\fbox{$a\ge \frac{3}{4}$}$$
See another answer here.
Ng Chung Tak
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