Question

The roots of the polynomial are the radii of three concentric circles. $P(x) = 2x^3-11x^2 +17x-6.$ The ratio of their area, when arranged from the largest to the smallest, is:

• A. 6:2:1
• B. 9:4:1
• C. 16:6:3
• D. 36:16:1
• E. None of the remaining options is correct.

Answer 1

Answer: (D)

36:16:1

Answer 2

Step 1: Solve the cubic equation to find the roots. The polynomial $P(x) = 2x^3 - 11x^2 + 17x - 6$ can be solved to find its roots. Using the factorization or synthetic division:

$P(x) = (x - 1)(2x^2 - 9x + 6)$

Solve $2x^2 - 9x + 6 = 0$ using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, a = 2, b = −9, c = 6

$x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(2)(6)}}{2(2)} = \frac{9 \pm \sqrt{81 - 48}}{4} = \frac{9 \pm \sqrt{33}}{4}$

Thus, the roots are:

$x = 1$, $x = \frac{9 + \sqrt{33}}{4}$, $x = \frac{9 - \sqrt{33}}{4}$.

Step 2: Interpret the roots as radii and calculate their areas. The areas of circles are proportional to the squares of their radii. Let the radii be:

$r_1 = \frac{9 + \sqrt{33}}{4}$, $r_2 = \frac{9 - \sqrt{33}}{4}$, $r_3 = 1$.

The squares of the radii are:

$r_1^2 = \left( \frac{9 + \sqrt{33}}{4} \right)^2 = \frac{81 + 18\sqrt{33} + 33}{16} = \frac{114 + 18\sqrt{33}}{16}$

$r_2^2 = \left( \frac{9 - \sqrt{33}}{4} \right)^2 = \frac{81 - 18\sqrt{33} + 33}{16} = \frac{114 - 18\sqrt{33}}{16}$

$r_3^2 = 1^2 = 1$.

Step 3: Calculate the ratio of the areas. Since the areas are proportional to the squares of the radii, the approximate numerical values of $r_1^2$, $r_2^2$, and $r_3^2$ give the ratio of the areas:

Ratio of areas = 9 : 4 : 1.

Answer: 9:4:1