Question

If $(1+2!)$ is a root of the equation $x^2+bx+c=0,$ where b and c are real then (b, c) is given by

• A. (2,-15)
• B. (-3, 1)
• C. (-2, 5)
• D. (3, 1)

Answer 1

The correct option is A (2,-15)

$\Rightarrow$ $x^2+bx+c=0$

1 root is $=(2!+1)=(2\times 1+1)=3$

putting in the equation we get

$\Rightarrow$ $3^2+3b+c=0$

$\Rightarrow$ $c=-9-3b.....\left(1\right)$

The equation (1) is ture for $(b,c)=(2,-15)$