The equation $x^2 + (k-3)x + (3-2k) = 0$, where $k$ is a constant, has two distinct real roots - Edexcel - A-Level Maths Pure - Question 9 - 2011 - Paper 2

To demonstrate that $k$ satisfies the inequality $k^2 + 2k - 3 > 0$, we first identify the roots of the equation.

The roots can be calculated using the quadratic formula: $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = 2$, and $c = -3$. Therefore, the calculation is:

= \frac{-2 \pm \sqrt{4 + 12}}{2} = \frac{-2 \pm \sqrt{16}}{2} = \frac{-2 \pm 4}{2}$$ This results in: 1. $k = \frac{2}{2} = 1$ 2. $k = \frac{-6}{2} = -3$ Next, we can factor the quadratic expression as follows: $$k^2 + 2k - 3 = (k - 1)(k + 3)$$ To satisfy the inequality $(k - 1)(k + 3) > 0$, we determine the intervals by testing values around the critical points $k = 1$ and $k = -3$. The intervals are: - $k < -3$ - $-3 < k < 1$ - $k > 1$ Testing values in these ranges, we find: 1. For $k < -3$, choose $k = -4$: $(-4-1)(-4+3) = (-5)(-1) = 5 > 0$ 2. For $-3 < k < 1$, choose $k = 0$: $(0-1)(0+3) = (-1)(3) = -3 < 0$ 3. For $k > 1$, choose $k = 2$: $(2-1)(2+3) = (1)(5) = 5 > 0$ Thus, $k$ satisfies the inequality $k^2 + 2k - 3 > 0$ when $k < -3$ or $k > 1$.