f(x) = x² + 4kx + (3 + 11k), where k is a constant - Edexcel - A-Level Maths Pure - Question 1 - 2009 - Paper 2

For the equation f(x) = 0 to have no real roots, the discriminant must be less than 0:

1. The general form of the quadratic equation is:

   $ax^2 + bx + c = 0$

   where:

   • a = 1
   • b = 4k
   • c = 3 + 11k

2. The discriminant (D) is given by:

   $D = b^2 - 4ac$

   Substituting the values of a, b, and c:

   $D = (4k)^2 - 4(1)(3 + 11k)$

   This simplifies to:

   $D = 16k^2 - (12 + 44k)$

   $D = 16k^2 - 44k - 12$

3. Set up the inequality for no real roots:

   $16k^2 - 44k - 12 < 0$

4. Solving this quadratic inequality will provide the ranges of k. First, solve the equation:

   $16k^2 - 44k - 12 = 0$

   Using the quadratic formula:

   $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{44 \pm \sqrt{(-44)^2 - 4(16)(-12)}}{2(16)}$

   Compute the roots to determine the intervals for k.

Given that k = 1, substitute this into f(x):

$f(x) = x^2 + 4(1)x + (3 + 11(1)) = x^2 + 4x + 14$

Next, complete the square:

$f(x) = (x + 2)^2 + 10$

1. The vertex of the graph is at (-2, 10). Since the parabola opens upwards and has a vertex above the x-axis, it will not cross the x-axis.

2. To find where it crosses the y-axis, set x = 0:

   $f(0) = (0 + 2)^2 + 10 = 4 + 10 = 14$

So the graph crosses the y-axis at (0, 14).

3. Sketch the graph, showing its shape, minimization at (−2, 10), and it crossing the y-axis at (0, 14).