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Let $k\in\mathbb{R}$ and $f(x)=x^{4}+(2-k^{2})x^{2}-2k^{2}x+(1-k^{2})$. Suppose that there are exactly two distinct real roots for the equation $f(x)=0$. Find the range of $k$.

I don't know the conditions that $f(x)=0$ has exactly two distinct real roots.

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Hint: express this polynomial to a difference of two polynomial squares. –  Yuchen Liu Mar 31 '13 at 6:41
    
Oh, I see. Thank you. –  user70096 Mar 31 '13 at 6:52
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1 Answer

$$f(x)=x^{4}+(2-k^{2})x^{2}-2k^{2}x+(1-k^{2})$$ $$f(x) = x^4+2x^2+1-k^2(x^2+2x+1)$$ $$f(x)=(x^2+1)^2 -k^2(x+1)^2$$ $$f(x)=(x^2+1)^2 -(k(x+1))^2$$ $$f(x)=((x^2+1)-k(x+1))((x^2+1)+k(x+1))$$ $$f(x)=(x^2+1-kx-k)(x^2+1+kx+k)$$

SO you have two polynomials $$x^2+1-kx-k\tag1$$ $$x^2+1+kx+k\tag2$$

Now Solving for x \begin{cases} - \frac{1}{2} k - \frac{1}{2} \sqrt{k^{2} - 4 k -4}\\ - \frac{1}{2} k + \frac{1}{2} \sqrt{k^{2} - 4 k -4}\\ \frac{1}{2} k - \frac{1}{2} \sqrt{k^{2} + 4 k -4}\\ \frac{1}{2} k + \frac{1}{2} \sqrt{k^{2} + 4 k -4}\\ \end{cases}

Now solve the inequality

\begin{cases} k^{2} - 4 k -4 \ge 0\\ k^{2} - 4 k -4 \ge 0\\ k^{2} + 4 k -4 \ge 0\\ k^{2} + 4 k -4 \ge 0\\ \end{cases}

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