方程 $3x^2 + 9x - k = 0$ 的两个根之差为 13 个单位，其中 $k$ 是常数。求 $k$ 的值。 答案 k = ...

1. Write the difference between the two roots in terms of the discriminant.

For $3x^2+9x-k=0$, $a=3$, $b=9$, $c=-k$.

The difference between the two roots is
$\frac{\sqrt{b^2-4ac}}{|a|}.$
Given that the difference is $13$,
$\frac{\sqrt{9^2-4(3)(-k)}}{3}=13.$

2. Simplify the equation involving the discriminant.

$\frac{\sqrt{81+12k}}{3}=13$
$\sqrt{81+12k}=39$

3. Square both sides and solve for $k$.

$81+12k=39^2$
$81+12k=1521$
$12k=1440$
$k=120$