Question

The quadratic equations $x^2-6x+a=0,x^2-cx+6=0$ have one root in common. The other roots of the first and second equations are integers in the ratio $4:3$. Then, the common root is

• A. $2$
• B. $1$
• C. $4$
• D. $3$

Answer 1

The correct option is A

$2$

Find the common roots

Let the common root be $\alpha$

As the other roots are in $4:3$

so,

Let $4\beta$ is the root of $x^2-6x+a=0$

and $3\beta$ is the root of $x^2-cx+6=0$

We know that

if $a{x}^2+bx+c$is the quadratic euqation, $p$and $q$be the roots of the equation.

then $p+q=-b/a$,$p\times q=c/a$

So, by this

$\alpha +4\beta =6$ and $4\alpha \beta =a$

$\alpha +3\beta =c$ and $3\alpha \beta =6$

as,

$$\begin{gathered} 3\alpha \beta =6 \\ \alpha \beta =6/3 \\ \alpha \beta =2 \end{gathered}$$

By this

$$\begin{gathered} 4\alpha \beta =a \\ 4\times 2=a \\ 8=a \end{gathered}$$

The first equation is $x^2-6x+8=0$

We can re write it as

$$\begin{gathered} x^2-2x-4x+8=0 \\ x(x-2)-4(x-2)=0 \\ (x-2)(x-4)=0 \end{gathered}$$

So,

$\begin{array}{rcl}x-2& =& 0\\ x& =& 2\end{array}$ and $\begin{array}{rcl}x-4& =& 0\\ x& =& 4\end{array}$

$x=2,4$

For $ɑ=2,4\beta =4$

so,

$\begin{array}{rcl}\beta & =& 4/4\\ & =& 1\end{array}$

$3\beta =3$

For $ɑ=4,4\beta =2$

so,

$\begin{array}{rcl}\beta & =& 2/4\\ & =& 1/2\end{array}$

$3\beta =3/2$

Hence the common root is $ɑ=2$.