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)

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

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

$$a+4c+4b=0$$