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$$ax^2 + bx + c = 0\quad\text{and}\quad bx^2 + cx + a = 0$$ 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?

nonuser
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BigV
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    Without further information, we cannot say that $1$ is a root. For example, if $a=b=c=1$, the two quadratic equations are $x^2+x+1=0$ and $x^2+x+1=0$. These clearly have a common root (as they are the same equation), but $1$ is not a root. From answers below though, we know that a common root $x_0$ must satisfy $$a\left(x_0^3-1\right)=0,$$ or equivalently (if $a\ne 0$) $$x_0^3=1.$$ Therefore, if you are given that there is a common real root, then you can conclude that $x_0=1$ is a common root. – Minus One-Twelfth Apr 06 '19 at 14:30
  • It's possible of course that this question came up in a context where the reader is not expected to know about complex numbers (just real numbers). In that case, when they said there is a common root, it automatically meant real root and you can more or less ignore my comment above. – Minus One-Twelfth Apr 06 '19 at 14:49
  • Could you clarify if the common root is required to be a real root? – Servaes Apr 06 '19 at 16:08

3 Answers3

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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 $x^2+x+1=0$.

nonuser
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1

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

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

farruhota
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