Question

Let $f\left(x\right)=a{x}^2+bx+c,a,bϵR,a\ne 0$ satisfying $f\left(1\right)+f\left(2\right)=0$ and and $-1$ is a root of the expression Then the equation $f\left(x\right)=0$ has

• A. No real root
• B. $1$ and $2$ are real roots
• C. Two distinct roots
• D. Two equal roots

Answer 1

Answer: (C)

Two distinct roots

$f\left(x\right)=a{x}^2+bx+c$ where $a\ne 0$

$f\left(1\right)+f\left(2\right)=0$

$\Rightarrow a{1}^2+b+c+a{2}^2+2b+c=0$

$\Rightarrow 5a+3b+2c=0$ ........$\left(1\right)$

$f(-1)=a-b+c=0$ ........$\left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$ we get

$\frac{a}{3+2}=\frac{b}{2-5}=\frac{c}{-5-3}=t$(say)

$\Rightarrow a=5t,b=-3t,c=-8t$

Substituting the above values in $a{x}^2+bx+c=0$

$8t{x}^2-3tx-8t=0$

$t\ne 0,5x^2-3x-8=0$

$5x^2+5x-8x-8=0$

$5x(x+1)-8(x+1)=0$

$(5x-8)(x+1)=0$

$x=\frac{8}{5},-1$

$\therefore x=\frac{8}{5}$ since $a>0$