$1$ as a common root of a quadratic equation

Question

$$ax^2 + bx + c = 0\quad\text{and}\quad bx^2 + cx + a = 0$$ have a common root.

In my book, it says that 1 is a common root for those equation?

Is this correct.

When we plug 1 in both the equations, we get $a+b+c = 0$, it says nothing about 1 being a root. Since we don't know if LHS is zero or not.

Where I'm going wrong here?

Answer 1

From first one you have $$c=-ax^2-bx$$ so put this in second eqaution and you get:$$ ax^3-a=0$$ Since $a\neq 0$ we have $x=1$ or solution of $x^2+x+1=0$.

Answer 2

Let $t$ is the common root,

$at^2+bt+c=0$

$bt^2+ct+a=0$

Solve for $t^2,t$

and use $t^2=(t)^2$ to eliminate $t$

and find $$0=a^3+b^3+c^3-3abc=(a+b+c)(?)$$

Answer 3

Let $t$ be the common root: $$at^2+bt+c=0\\ bt^2+ct+a=0$$ Subtract and factorize: $$a(t^2-1)+b(t-t^2)+c(1-t)=0 \iff \\ a(t-1)(t+1)-bt(t-1)-c(t-1)=0 \iff \\ (t-1)(at+a-bt-c)=0 \Rightarrow t=1$$