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 10 - 2011 - Paper 2

To solve the inequality $k^2 + 2k - 3 > 0$, we first find the roots of the corresponding equation:

$k^2 + 2k - 3 = 0$

Applying the quadratic formula, where $a = 1$, $b = 2$, and $c = -3$, we have:

$k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-2 \pm \sqrt{(2)^2 - 4(1)(-3)}}{2(1)}$

Calculating the discriminant:

$= \frac{-2 \pm \sqrt{4 + 12}}{2} = \frac{-2 \pm \sqrt{16}}{2} = \frac{-2 \pm 4}{2}$

The roots are:

$k_1 = \frac{2}{2} = 1$ $k_2 = \frac{-6}{2} = -3$

Next, we can examine the sign of $k^2 + 2k - 3$ in the intervals determined by these roots: $(-\infty, -3)$, $(-3, 1)$, and $(1, \infty)$.

• For $k < -3$: Choose $k = -4$: $(-4)^2 + 2(-4) - 3 = 16 - 8 - 3 = 5 > 0$

• For $-3 < k < 1$: Choose $k = 0$: $0^2 + 2(0) - 3 = -3 < 0$

• For $k > 1$: Choose $k = 2$: $2^2 + 2(2) - 3 = 4 + 4 - 3 = 5 > 0$

Thus, the solution to the inequality $k^2 + 2k - 3 > 0$ is:

$k < -3 \quad \text{or} \quad k > 1.$

Therefore, the set of possible values of $k$ is:

$k \in (-\infty, -3) \cup (1, \infty).$