Question

If $x^3-px^2+qx-27=0$ has three positive roots where p, q>0 and $p+k+l^2=4l+10$ where $k, l \in \mathbb{R}$, then the greatest value of k is

• A. 1
• B. 5
• C. 6
• D. 7

Answer 1

Answer: (B)

5

Answer 2

Let the three positive roots of the cubic equation $x^3-px^2+qx-27=0$ be $\alpha, \beta, \gamma$. By Vieta's formulas:

1. Sum of the roots: $\alpha + \beta + \gamma = p$
2. Product of the roots: $\alpha\beta\gamma = 27$

Since the roots are positive, we can apply the AM-GM inequality: $\frac{\alpha + \beta + \gamma}{3} \ge \sqrt[3]{\alpha\beta\gamma}$ Substituting the values: $\frac{p}{3} \ge \sqrt[3]{27}$ $\frac{p}{3} \ge 3$ $p \ge 9$ The minimum value of $p$ is 9.

We are given $p+k+l^2=4l+10$. Rearranging for $k$: $k = 4l + 10 - l^2 - p$ $k = 10 - p + (4l - l^2)$ To maximize $k$, we need to maximize the term $(4l - l^2)$ and use the minimum value of $p$. The expression $4l - l^2$ can be maximized by completing the square: $4l - l^2 = -(l^2 - 4l) = -( (l-2)^2 - 4 ) = 4 - (l-2)^2$ The maximum value of $4l - l^2$ is 4, which occurs when $l=2$.

Now, substitute the maximum value of $4l-l^2$ and the minimum value of $p$ into the expression for $k$: $k_{max} = 10 - p_{min} + (4l - l^2)_{max}$ $k_{max} = 10 - 9 + 4$ $k_{max} = 5$ The greatest value of $k$ is 5.