Mathematico | Mathematico

The vertex of the parabola

y = x^{2} + b x + 5

lies upon the straight line

x + y = - 1

Let the two possible values of $b$ be $b_{1}$ and $b_{2}$ .

Find the value of

b_{1} b_{2}^{2} + b_{2} b_{1}^{2}

Hint

You should begin completing the square by writing

{(x + \frac{b}{2})}^{2} - \dots

Hint

The $x$ and $y$ coordinate of the vertex satisfy

x + y = - 1

Solution

In complete square form,

\begin{aligned} y & = {(x + \frac{b}{2})}^{2} - \frac{b^{2}}{4} + 5 \\ = {(x + \frac{b}{2})}^{2} + \frac{20 - b^{2}}{4} \end{aligned}

and so the vertex has coordinates

(- \frac{b}{2}, \frac{20 - b^{2}}{4})

These coordinates satisfy $x + y = - 1$ , so

\begin{aligned} - \frac{b}{2} + \frac{20 - b^{2}}{4} & = - 1 \\ - 2 b + 20 - b^{2} & = - 4 \\ b^{2} + 2 b - 24 & = 0 \\ (b + 6) (b - 4) & = 0 \\ b & \in {- 6, 4} \end{aligned}