Question

If $a,b\in R,a\ne 0$ and quadratic equation $a{x}^2-bx+1=0$ has imaginary roots, then $(a+b+1)$ is

• A. Positive
• B. Negative
• C. Zero
• D. Dependent on the sign of $b$

Answer 1

The correct option is A Positive

${ax}^2-bx+1=0hasimaginaryroots\therefore D<0b^2-4a<0b^2<4ale{b}^2<4a\therefore a>0\therefore f\left(x\right)>0for\forall x\in R\therefore f(-1)=a+b+1>0$