Question

The number of values of the pair $(a,b)$ for which $a(x+1{)}^2+b(x^2-3x-2)+x+1=0$ is an identity in $x$ is

• A. $0$
• B. $1$
• C. $2$
• D. Infinite

Answer 1

Answer: (B)

$1$

$a{\left(x+1\right)}^2+b\left(x^2-3x-2\right)+x+1=0$
$a\left(x^2+2x+1\right)-b\left(x^2+3x+2\right)+x+1=0$
$a{x}^2+ax+ax+a-b[\left(x+1\right)x+2x+2]+x+1=0$
$ax\left(x+1\right)+a\left(x+1\right)-b\left(x+1\right)\left(x+2\right)+\left(x+1\right)=0$
$\left(x+1\right)\left[ax+a-b\left(x+2\right)+1\right]=0$
$ax+a-bx-2b+1=0$
$x\left(a-b\right)+\left(a-2b+1\right)=0$
$\therefore a-b=0anda-2b+1=0$
$a=b$
$a-2a=-1$
$\therefore a=1=b$