Given the expression: y = x² + 2x + 3 = (x + a)² + b - Edexcel - A-Level Maths Pure - Question 1 - 2005 - Paper 1

To sketch the graph:

1. Identify the vertex as calculated: ((–1, 2))
2. The y-intercept can be found by setting $x = 0$: $y = 0^2 + 2(0) + 3 = 3 \Rightarrow (0, 3)$
3. To find the x-intercepts, set $y = 0$: $0 = x^2 + 2x + 3$ The discriminant is negative, indicating no real x-intercepts.

The graph is a 'U'-shaped parabola opening upwards with vertex at ((–1, 2)) and passing through the y-axis at ((0, 3)).

The discriminant $D$ of a quadratic equation $ax^2 + bx + c = 0$ is given by: $D = b^2 - 4ac$ For the equation $x^2 + 2x + 3$:

• Here, $a = 1$, $b = 2$, and $c = 3$.
• Substitute values into the discriminant: $D = (2)^2 - 4(1)(3) = 4 - 12 = -8$

Since the discriminant is negative ($D < 0$), it confirms there are no real roots, aligning with our graph sketch, which shows the parabola does not intersect the x-axis.

For the quadratic equation $x^2 + kx + 3 = 0$ to have no real roots, the discriminant must be negative: $D = k^2 - 4(1)(3) < 0$ This simplifies to: $k^2 - 12 < 0$ Thus: $k^2 < 12$ This gives: $-\sqrt{12} < k < \sqrt{12}$ In surd form: $-2\sqrt{3} < k < 2\sqrt{3}$