Mathematico | Mathematico

The roots of the quadratic

x^{2} + (k + 2) x + k, k < 0

are $α$ and $β$ .

Given that $β$ is $3$ larger than $α$ , find the value of

\frac{β^{2}}{α^{2}}

Hint

We have

\begin{aligned} a & = 1 \\ b & = (k + 2) \\ c & = k \end{aligned}

Hint

The larger root minus the smaller root is equal to $3$ .

Solution

The formula gives the roots as

\begin{aligned} x & = \frac{- (k + 2) \pm \sqrt{(k + 2)^{2} - 4 k}}{2} \\ = \frac{- (k + 2) \pm \sqrt{k^{2} + 4}}{2} \end{aligned}

The larger root minus the smaller root is equal to $3$ , so

\begin{aligned} \frac{- (k + 2) + \sqrt{k^{2} + 4}}{2} - \frac{- (k + 2) - \sqrt{k^{2} + 4}}{2} & = 3 \\ \sqrt{k^{2} + 4} & = 3 \\ k^{2} + 4 & = 9 \\ k & = \pm \sqrt{5} \end{aligned}

But $k < 0$ , so $k = - \sqrt{5}$ .

To find the roots, we must solve

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

We of course use the formula

\begin{aligned} x & = \frac{- (2 - \sqrt{5}) \pm \sqrt{(2 - \sqrt{5})^{2} - 4 (1) (- \sqrt{5})}}{2} \\ x & = \frac{\sqrt{5} - 2 \pm 3}{2} \\ x & \in {\frac{\sqrt{5} - 5}{2}, \frac{\sqrt{5} - 1}{2}} \end{aligned}