Question

If $x^2+mx+n=0$ and $x^2+px+q=0$ have a common root, then the common root is

• A. $\frac{q-n}{m-p}$
• B. $\frac{q-n}{m+p}$
• C. $\frac{q+n}{m+p}$
• D. None of these

Answer 1

The correct option is A

$\frac{q-n}{m-p}$

Explanation for the correct option:

Find the common root:

Let $\alpha$ be the common root of the equations $x^2+mx+n=0$ and $x^2+px+q=0$. Then $\alpha$ satisfies both the equations.

$$\begin{gathered} {\alpha }^2+m\alpha +n=0...\left(1\right) \\ {\alpha }^2+p\alpha +q=0...\left(2\right) \end{gathered}$$

By the rule of cross multiplication, the solution of two simultaneous equations:

$$\begin{gathered} a_{1}x+b_{1}y+c_1=0 \\ {a}_{2}x+b_{2}y+c_2=0\mathrm{is} \\ \frac{\mathrm{x}}{{\mathrm{b}}_1{\mathrm{c}}_2-{\mathrm{b}}_2{\mathrm{c}}_1}=\frac{\mathrm{y}}{{\mathrm{c}}_1{\mathrm{a}}_2-{\mathrm{c}}_2{\mathrm{a}}_1}=\frac{1}{{\mathrm{a}}_1{\mathrm{b}}_2-{\mathrm{a}}_2{\mathrm{b}}_1}...\left(3\right) \end{gathered}$$

By solving $\left(1\right)$ and $\left(2\right)$ simultaneously, we get

$\frac{{\alpha }^2}{mq-np}=\frac{\alpha }{n-q}=\frac{1}{p-m}\left[\mathrm{from}\left(3\right)\right]$

From the last two term ,we get

$$\begin{gathered} \Rightarrow \frac{\alpha }{n-q}=\frac{1}{p-m} \\ \Rightarrow \alpha =\frac{n-q}{p-m} \\ \Rightarrow \alpha =\frac{-\left(q-n\right)}{-\left(m-p\right)} \\ \Rightarrow \alpha =\frac{\left(q-n\right)}{\left(m-p\right)} \end{gathered}$$

Hence, the correct option is A.