Make the roots exceed 3

Algebra Level 3

Find the least positive integer value of $m$ for which both roots of the equation

$x^{2} - 6mx + 9m^{2} - 2m + 2 = 0$

are greater than or equal to 3.

The answer is 2.

1 solution

Chew-Seong Cheong
Dec 16, 2014

The roots of the equation are given by:

$\quad x = \dfrac {-(-6m) \pm \sqrt{(-6m)^2-4(1)(9m^2-2m+2}}{2(1)}$

$\quad \quad = \dfrac {6m \pm \sqrt{36m^2-36m^2+8m-8}}{2} = 3m \pm \sqrt{2(m-1)}$

For both roots to exceed 3, for $m > 0$ , we have:

$\Rightarrow 3m - \sqrt{2(m-1)} > 3\quad \Rightarrow 3(m-1) > \sqrt{2(m-1)}$

$\Rightarrow 9(m-1)^2 > 2(m-1) \quad \Rightarrow 9(m-1) > 2$

$\Rightarrow 9m > 11 \quad \Rightarrow m > \dfrac {11}{9} \quad \Rightarrow m = \boxed {2}$

I'm confused by this solution. The problem asks us for value of $m$ for which both (not necessarily distinct) roots of the equation are greater or equal to 3. Choosing $m = 1$ makes the equation to have double root $x = 3$ . Is there a problem in my thinking or is this solution incorrect?

Jan Hrček - 5 years, 11 months ago

The problem asks for both roots $x_1$ and $x_2$ greater or equal to 3, that is $x_1 \ge 3$ and $x_2 \ge 3$ and NOT $x_1=x_2 \ge 3$ .

Chew-Seong Cheong - 5 years, 11 months ago

Make the roots exceed 3

The answer is 2.

1 solution

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