Are there any two real nonzero numbers $a$ and $b$ such that $a^2+ab+b^2 = 0$ Why or why not?

Question

Are there any two real nonzero numbers $a$ and $b$ such that $a^2+ab+b^2 = 0$ Why or why not?

$$a^2+ab+b^2=(a+b)^2-ab=0$$ iff $$(a+b)^2=ab \tag{1}$$ but $(a+b)^2 = a^2+2ab+b^2 $ so equation 1 couldn't possibly be true.

Also, when $a=b\ne 0$, $(a^2+ab+b^2)(a-b) = a^3-b^3 =0$.

Answer 1

Hint :

$$a^2 + ab + b^2 = \left( a + \frac{b}{2}\right)^2 + \frac{3b^2}{4}$$

Answer 2

Fix $b\neq0$ and try to solve in $a$. This a second order equations so the possible values of $a$ in terms of $b$ are

$$ a_{1,2}=\frac{-b\mp \sqrt{-3b^2}}{2}$$

So, for any non zero real value $b$ the possible values of $a$ are complex conjugated because $-3b^2<0$. Therefore you cannot find two non-zero reals such that $(a+b)^2=ab$.

Answer 3

One way is to start from $$a^2+ab+b^2=b^2\left(\left(\frac{a}{b}\right)^2+\frac{a}{b}+1\right)=0$$ so letting $x=\frac{a}{b}$ we get the quadratic equation $x^2+x+1=0$ which has no real roots. This is similar to the idea hinted at in the comment by RiverX15.

Answer 4

$$a^2 + ab + b^2 = \left( a + \frac{b}{2}\right)^2 + \frac{3b^2}{4}=0$$ Since the two terms are non-negative, thus $$\left( a + \frac{b}{2}\right)^2= \frac{3b^2}{4}=0$$ Therefore $$a=0,\quad b=0$$