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Given that $a$ and $b$ are real constants and that the equation $x^4+ax^3+2x^2+bx+1=0$ has at least one real root, find the minimum possible value of $a^2+b^2$.

I began this way: Let the polynomial be factorized as $(x^2+\alpha x + 1)(x^2+\beta x +1)$. Then expanding and comparing coefficients we get $\alpha\beta=0$, meaning either $\alpha=0$ or $\beta=0$. Suppose $\alpha=0$. Then we see that $(x^2+\beta x+1)$ should have real roots, from which we get $\beta^2 \geq 4$. But we get $a=b=\beta$ from the comparison above. So $a^2+b^2 = 2\beta^2 \geq 8$.

Is it correct? Or is there any mistake? Any other solution is also welcome.

Yellow
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    How do you know the polynomial has such a factorisation? – Angina Seng Jan 10 '19 at 05:27
  • Any fourth degree polynomial can be factorised into product of two quadratic expressions, no matter what the nature of its roots are, right? – Yellow Jan 10 '19 at 05:32
  • @Any Radha, yes but not in that way you write... – dmtri Jan 10 '19 at 05:38
  • I can't find anything wrong in your solution. – Anik Bhowmick Jan 10 '19 at 05:43
  • Well, @AnikBhowmick, I was not sure of the constant coefficients in the factorisation. – Yellow Jan 10 '19 at 05:44
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    Wouldn't this imply that $\alpha + \beta = a = b$ which need not be the case? If $x^4 + ax^3 + 2x^2 +bx +1 = (m x^2 + nx + p)(jx^2 + kx + q)$ we know $j = \frac 1m$ and wolog it factors to $(x^2 + \alpha x + \gamma)(x^2 + \beta x + \delta)$ but we can't assume $\gamma = \delta = 1$ – fleablood Jan 10 '19 at 06:24
  • .... which gives as $\delta + \gamma + \alpha\beta = 2$. But $\delta + \gamma = \delta + \frac 1\delta \ge 2$(if we assume $\delta > 0$ which wolog I think we can). So $\alpha\beta \le 0$ ... not sure we can assume that $\alpha$ and $\beta$ are both the same signage.... – fleablood Jan 10 '19 at 06:32

2 Answers2

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$$x^4+ax^3+2x^2+bx+1=$$ $$=(x^2+\frac{a}{2}x)^2-\frac{a^2}{4}x^2+2x^2+bx+1=$$ $$=(x^2+\frac{a}{2}x)^2-\frac{a^2}{4}x^2+2x^2-\frac{b^2}{4}x^2+(\frac{b}{2}x+1)^2=$$ $$=(x^2+\frac{a}{2}x)^2+x^2(2-\frac{a^2+b^2}{4})+(\frac{b}{2}x+1)^2$$

If $2-\frac{a^2+b^2}{4}>0$, then we have a problem; the sum of three nonnegative terms equals zero (at our real root), which can only happen if $x=0$ to make the second term zero. But then $(\frac{b}{2}x+1)^2>0$. So if $a^2+b^2<8$ there is no real root. Hence $a^2+b^2\ge 8$.

Now, if $a^2+b^2=8$, then the middle term is zero. Can we make the other two terms zero as well? We need $x=-\frac{a}{2}$ and $x=-\frac{2}{b}$, to make the first and last terms zero. This can be achieved, with $a=b=2$ and $x=-1$.

Hence, the minimum possible value of $a^2+b^2$ is indeed $8$.

vadim123
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I would write first $(x^2+cx+d) (x^2+ex+f) $ and then find $c, d, e, f$ in terms of $a, b$.

dmtri
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