# How can the sum of squares be negative?

If $a,b,c,d$ are the roots of the equation $x^4-Kx^3+Kx^2+Lx+M=0$, where $K,L,M$ are real numbers, then the mininmum value of $a^2+b^2+c^2+d^2$ is?

My answer:

$\sum a=K,\ \sum ab=K\implies$

$a^2+b^2+c^2+d^2=K^2-2K=(K-1)^2-1$. For $K=1$, $(a^2+b^2+c^2+d^2)_{min}=-1$

This matches with the answer in fact, but how can sum of squares ever result in NEGATIVE.

What's the intuition behind this answer is it wrong or I'm going the wrong way.

• Hint: The coefficients are real, but what about the roots? – Ronald Blaak Jul 10 '17 at 13:27
• Are you aware about complex numbers? – Tob Ernack Jul 10 '17 at 13:28
• Complex roots... – MCCCS Jul 10 '17 at 13:28
• @TobErnack yes a bit of. – mnulb Jul 10 '17 at 13:30
• The sum of the squares of the roots of $x^2+1=0$ is also negative. – G Tony Jacobs Jul 10 '17 at 13:32

## 2 Answers

Consider $K=1,$ that is, $$f(x)=x^4-x^3+x^2+Lx+M.$$ We have that

$$f''(x)=2(6x^2-3x+1).$$ Note that $$f''=0$$ has no real roots. Thus $f'=0$ has a real root and so $f=0$ has at most two real roots. In other word, at least two roots of the equation $$f(x)=0$$ are complex. Thus there is no contradiction with the fact that the sum of the squares of the roots is $-1.$

If $L=M=0$ then the roots are $0$ (double) and $\dfrac{1\pm\sqrt {-3}}{2}=\dfrac{1\pm i\sqrt{3}}{2}.$ We have that

$$0^2+0^2+\left(\dfrac{1+i\sqrt{3}}{2}\right)^2+\left(\dfrac{1-i\sqrt{3}}{2}\right)^2=-1.$$

• $f''=0$ has no real roots, I'm not aware of this can you explain why? – mnulb Jul 11 '17 at 3:15
• Because $6x^2+3x+1=0\iff x=\frac{3\pm\sqrt{\color{red}{9-24}}}{12}$ – mfl Jul 11 '17 at 8:26

Hint:

the roots of the quartic can be complex numbers and in this case we have couples of conjugate roots of the form $\alpha \pm i \beta$ and: $$(\alpha+i\beta)^2 +(\alpha-i\beta)^2=2\alpha^2-2\beta^2$$