Question

If $x^2-11x+a=0$ and $x^2-14x+2a=0$ have one common root then a is equal to -

• A. $0,-24$
• B. $0,1$
• C. $0,24$
• D. $1,24$

Answer 1

The correct option is C $0,24$

Consider the equations

$x^2-11x+a=\mathrm{0...}\left(1\right)$

and

$x^2-14x+2a=\mathrm{0...}\left(2\right)$

Let $\alpha ,\beta$ are the roots of equation (1)

$\therefore$ Sum of roots of equation (1) is

$\alpha +\beta =-\frac{B}{A}=-\frac{(-11)}{1}=\mathrm{11...}\left(3\right)$

Products of roots of equation (1) is

$\alpha \beta =\frac{C}{A}=\frac{a}{1}=a...\left(4\right)$

Let $\alpha ,\gamma$ are the roots of equation (2)

$\therefore$ Sum of roots of equation (2) is

$\alpha +\gamma =\frac{B}{A}=\frac{(-14)}{1}=\mathrm{14...}\left(5\right)$

Products of roots of equation (2) is

$\alpha \gamma =\frac{C}{A}=\frac{2a}{1}=2a...\left(6\right)$

Using equations, (3), (4), (5), (6) solve for $\alpha ,\beta ,\gamma .$

Thus, we have,

$\alpha =8,\beta =3,\gamma =6$

Substituting the values of $\alpha$ and $\beta$ in equation (4), we have

$a=\alpha \beta \Rightarrow a=8\times 6=24$