# 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?

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

$a^2+b^2+c^2+d^2=K^2-2K=(K-1)^2-1$. 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? Commented Jul 10, 2017 at 13:27
• Are you aware about complex numbers? Commented Jul 10, 2017 at 13:28
• Complex roots... Commented Jul 10, 2017 at 13:28
• @TobErnack yes a bit of. Commented Jul 10, 2017 at 13:30
• The sum of the squares of the roots of $x^2+1=0$ is also negative. Commented Jul 10, 2017 at 13:32

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? Commented Jul 11, 2017 at 3:15
• Because $6x^2+3x+1=0\iff x=\frac{3\pm\sqrt{\color{red}{9-24}}}{12}$
– mfl
Commented Jul 11, 2017 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: $$(\alpha+i\beta)^2 +(\alpha-i\beta)^2=2\alpha^2-2\beta^2$$