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The parabola drawn below has a $y$ intercept of $4$ . The roots occur at $x = α$ and $x = β$ where $α < 0 < β$ . The roots and the $y$ intercept form a triangle of area $12$ ,

Parabola

Given that the parabola has the form

y = a x^{2} - a x + b

Find the value of $80 (α^{3} + β^{3})$ .

Hint

To find $b$ , look at the $y$ intercept, noting that $x = 0$ at this point.

Hint

Consider the area of a triangle to work out the value of $β - α$ .

Hint

Since $α, β$ are roots of the quadratic, we know that

\begin{aligned} a β^{2} - a β + 4 & = 0 \\ a α^{2} - a α + 4 & = 0 \end{aligned}

Try subtracting these equations from each other.

Solution

By considering the $y$ intercept, we know that $b = 4$ .

$β - α$ represents the base of the given triangle. The area is $12$ and the height is $4$ , so

\begin{aligned} \frac{1}{2} \times (β - α) \times 4 & = 12 \\ β - α & = 6 \end{aligned}

We know that $α$ and $β$ are the roots, so

\begin{aligned} a β^{2} - a β + 4 & = 0 \\ a α^{2} - a α + 4 & = 0 \end{aligned}

Subtracting these equations gives

\begin{aligned} a β^{2} - a β - a α^{2} + a α & = 0 \\ β^{2} - α^{2} - β + α & = 0 \\ (β - α) (β + α) - (β - α) & = 0 \\ 6 (β + α) - 6 & = 0 \\ β + α & = 1 \end{aligned}

We have shown that

\begin{aligned} β - α & = 6 \\ β + α & = 1 \end{aligned}

Adding the equations, we get

\begin{aligned} 2 β & = 7 \\ β & = \frac{7}{2} \end{aligned}

Subtracting them, we get

\begin{aligned} - 2 α & = 5 \\ α & = - \frac{5}{2} \end{aligned}

In the end, $80 (α^{3} + β^{3}) = 2180$ .