Question

If the quadratic equation $a{x}^2+2cx+b=0$ and $a{x}^2+2bx+c=0(b\ne c)$ have a common root, then $a+4b+4c=$

• A. $-2$
• B. $-1$
• C. $0$
• D. $1$

Answer 1

The correct option is C $0$

Consider equation 1
$a{x}^2+2cx+b=0$ ...(i)
Similarly
$a{x}^2+2bx+c=0$ ...(ii)
Subtracting ii from i
$2x(c-b)+(b-c)=0$
$(c-b)[2x-1]=0$
Let the common root be $\alpha$.
Hence
$(c-b)[2\alpha -1]=0$
$\alpha =\frac{1}{2}$
Substituting in equation i
$\frac{a}{4}+\frac{2c}{2}+b=0$
$\frac{a}{4}+c+b=0$
$a+4c+4b=0$
Hence
$a+4b+4c=0$.