Given the equation $$x^2 + 2x + 3 = (x + a)^2 + b.$$ (a) Find the values of the constants a and b - Edexcel - A-Level Maths Pure - Question 2 - 2005 - Paper 1

To sketch the parabola described by the function $y = x^2 + 2x + 3$, we start by finding its vertex. Using the vertex formula for the quadratic $y = ax^2 + bx + c$, the x-coordinate of the vertex is given by:

$x = -\frac{b}{2a} = -\frac{2}{2 \cdot 1} = -1.$

Substituting $x = -1$ into the equation to find the y-coordinate:

$y = (-1)^2 + 2(-1) + 3 = 1 - 2 + 3 = 2.$

Thus the vertex is at $(-1, 2)$.

To find the intercepts, we solve for when $y = 0$:

$x^2 + 2x + 3 = 0.$

Calculating the discriminant gives:

$D = b^2 - 4ac = 2^2 - 4 \cdot 1 \cdot 3 = 4 - 12 = -8,$

indicating there are no x-intercepts. The y-intercept occurs at $x = 0$:

$y = 0^2 + 2(0) + 3 = 3,$ giving the intercept (0, 3).

Based on this, the graph is a 'U'-shaped parabola opening upwards, with the vertex at (-1, 2) and a y-intercept at (0, 3), and no x-intercepts.

The discriminant of the quadratic equation $ax^2 + bx + c$ is given by:

$D = b^2 - 4ac.$

For our quadratic $x^2 + 2x + 3$, we have:

$D = 2^2 - 4 \cdot 1 \cdot 3 = 4 - 12 = -8.$

Since the discriminant is negative ($D < 0$), this indicates that there are no real roots for the quadratic equation, which means the graph does not intersect the x-axis. This is consistent with our sketch in part (b), where we observe that the parabola lies entirely above the x-axis.

To find the values of k for which the equation $x^2 + kx + 3 = 0$ has no real roots, we again use the discriminant, which must be less than zero:

$D = k^2 - 4 \cdot 1 \cdot 3 < 0$

This simplifies to:

$k^2 - 12 < 0 \Rightarrow k^2 < 12$

Taking the square root gives:

$-\sqrt{12} < k < \sqrt{12}$

Simplifying, we find:

$-2\sqrt{3} < k < 2\sqrt{3}.$

Thus, the set of possible values of k is $(-2\sqrt{3}, 2\sqrt{3}).$