Question

If the equation $x^3-a{x}^2+bx-a=0$ has three real roots then which of the following is true (CAT 2000)

• A. a = 11
• B. a ≠ 1
• C. b = 1
• D. b ≠ 1

Answer 1

The correct option is D b ≠ 1

Let $f\left(x\right)=x^3-a{x}^2+bx-a=0$

In the given equation, there are 3 sign changes, therefore there are at most 3 positive roots (Descarte’s Rule). If f(-x), there is no sign change. Thus there is no negative real root, i.e., if $\alpha ,\beta$ and $\gamma$ are the roots then they are all positive and we have

$f\left(x\right)=(x-\alpha )(x-\beta )(x-\gamma )=0$

$\to b=\alpha \beta +\beta \gamma +\gamma \alpha \to a=\alpha +\beta +\gamma =\alpha \beta \gamma$

$\to \frac{\alpha +\beta +\gamma }{\alpha \beta \gamma }=1\to \frac{1}{\alpha \beta }+\frac{1}{\beta \gamma }+\frac{1}{\gamma \alpha }=1$

$\to \alpha \beta ,\beta \gamma ,\gamma \alpha >1\to b>3.$

Thus $b\ne 1.$