If the quadratic equation, $ax^2+2cx+b=0$ & $ax^{2}+2bx-5x+c=0 (b \neq c)$ hava a common root then find the value of $a+4b+4c$.

Solution

Considering the equations to be:

$ax^2+2cx+b=0$...(1)

$ax^2+2bx+c=0$...(2)

(2)-(1) gives

$2(b-c)x+c-b=0$

$2(b-c)x=b-c$

$2x=1$

$\therefore x=\dfrac{1}{2}$

From (1)

$ax^2+2cx+b=0$

$a\left (\dfrac{1}{2} \right )^2+2c\dfrac{1}{2}+b=0$

$a+4c+4b=0$