Mathematico | Mathematico

One of the roots of the quadratic equation

x^{2} - 2 m x + (2 m + 5) = 0, m > 0

is twice as large as the other.

Given that both roots are positive, find their product as an exact decimal.

Hint

Completing the square gives you

(x - m)^{2} - m^{2} + 2 m + 5 = 0

so that

x = m \pm \sqrt{m^{2} - 2 m - 5}

The larger of these roots is twice the smaller.

Hint

Use the previous hint to find $m$ , then go back to solve the original equation.

Solution

Completing the square, we get

\begin{aligned} (x - m)^{2} - m^{2} + 2 m + 5 & = 0 \\ x & = m \pm \sqrt{m^{2} - 2 m - 5} \end{aligned}

The larger root is twice the smaller, so

\begin{aligned} m + \sqrt{m^{2} - 2 m - 5} & = 2 (m - \sqrt{m^{2} - 2 m - 5}) \\ m & = 3 \sqrt{m^{2} - 2 m - 5} \\ m^{2} & = 9 (m^{2} - 2 m - 5) \\ 8 m^{2} - 18 m - 45 & = 0 \\ (4 m - 15) (2 m + 3) & = 0 \\ m & = \frac{15}{4} \end{aligned}

The original equation becomes

\begin{aligned} x^{2} - \frac{15}{2} x + \frac{25}{2} & = 0 \\ 2 x^{2} - 15 x + 25 & = 0 \\ (2 x - 5) (x - 5) & = 0 \\ x & \in {2.5, 5} \end{aligned}

So the product of the roots is $12.5$ .