Question

Let $b$ be a non-zero real number. Suppose the quadratic equation $2x^2+bx+\frac{1}{b}=0$ has two distinct real roots. Then

• A. $b+\frac{1}{b}>\frac{5}{2}$
• B. $b+\frac{1}{b}<\frac{5}{2}$
• C. $b^2+\frac{1}{b^2}<4$
• D. $b^2-2b>0$

Answer 1

The correct option is D $b^2-2b>0$

If the equation $2x^2+bx+\frac{1}{b}=0$ has two distinct real roots, then $\Delta >0$

$\Rightarrow b^2-\frac{8}{b}>0\Rightarrow \frac{b^3-8}{b}>0\Rightarrow \frac{(b-2)(b^2+2b+4)}{b}>0\text{As we know}{b}^2+2b+4=(b+1{)}^2+3>0\text{for all real}b\Rightarrow \frac{(b-2)}{b}>0\Rightarrow b\in (-\infty ,0)\cup (2,\infty )$

Which is same as the solution of $b^2-2b>0$