Mathematico | Mathematico

The parabola

y = 4 x^{2} + 4 (2 k + 1) x + 2 k^{2} + 4 k + 9

is sketched below.

Parabola with one x intercept

Find the sum of all possible values of $k$ .

Hint

There's only one root, so $Δ = 0$ .

Only one of the values of $k$ you get is valid. Can you decide which one?

Hint

Notice that the root occurs when $x$ is positive - this allows you to eliminate one value of $k$ .

Solution

The diagram tells us there is only one root, so

\begin{aligned} b^{2} - 4 a c & = 0 \\ 16 (2 k + 1)^{2} - 4 (4) (2 k^{2} + 4 k + 9) & = 0 \\ (2 k + 1)^{2} - 2 k^{2} - 4 k - 9 & = 0 \\ 4 k^{2} + 4 k + 1 - 2 k^{2} - 4 k - 9 & = 0 \\ 2 k^{2} - 8 & = 0 \\ k & = \pm 2 \end{aligned}

When $k = 2$ , the root is given by

\begin{aligned} 4 x^{2} + 4 (2 k + 1) x + 2 k^{2} + 4 k + 9 & = 0 \\ 4 x^{2} + 20 x + 25 & = 0 \\ x & = \frac{- 5}{2} \end{aligned}

This does not make sense with respect to the diagram, since the root shown is positive.

However, when $k = - 2$ , the root is given by

\begin{aligned} 4 x^{2} + 4 (2 k + 1) x + 2 k^{2} + 4 k + 9 & = 0 \\ 4 x^{2} - 12 x + 9 & = 0 \\ x & = \frac{3}{4} \end{aligned}

This does make sense.

So the only possible value of $k$ is $k = - 2$ .