Conditions on polynomials with common roots.

Question

If one root of the equation $x^2 + ax + b = 0$ and $x^2 + bx + a = 0$ is common and $a \ne b$ then:

The options are as follows: $$\begin{array}{ll} (A)\quad& a + b = 0\\ (B)& a + b = -1\\ (C)& a - b = 1\\ (D)& a + b = 1 \end{array}$$

Idk how to solve this, please help me.

You should show what you've tried. – Macavity Oct 29 '14 at 18:46 — Macavity, Oct 29 '14 at 18:46

score 1 · Answer 1 · answered Oct 29 '14 at 18:47

1

Hint: Substitute the common root and subtract.

answered Oct 29 '14 at 18:47

Macavity

score 0 · Answer 2 · answered Oct 29 '14 at 18:51

0

$x^2+ax+b=x^2+bx+a$ => $(a-b)(x-1)=0$. Since $a \neq b$, $x=1$. Thus $1^2+a+b=0$. Hence, $a+b=-1$.

Thus, C is the answer.

answered Oct 29 '14 at 18:51

bankrip

score 0 · Accepted Answer · answered Oct 29 '14 at 18:53

0

Let $c$ be the common root. Then $$c^2+ac+b=0$$ $$c^2+bc+a=0$$ so $ac+b=bc+a$ or $(a-b)c=a-b$. Since $a\neq b$, $c$ must be $1$.

Now we have $a+b=-1$.

answered Oct 29 '14 at 18:53

ajotatxe

3 Answers3