A-Level Edexcel Maths Pure Polynomials: The equation $x^2 + kx + (k + 3)

The equation $x^2 + kx + (k + 3) = 0$, where $k$ is a constant, has different real roots - Edexcel - A-Level Maths Pure - Question 9 - 2007 - Paper 1

Question 9

The equation $x^2 + kx + (k + 3) = 0$, where $k$ is a constant, has different real roots. (a) Show that $k^2 - 4k - 12 > 0$. (b) Find the set of possible values of... show full transcript

Worked Solution & Example Answer:The equation $x^2 + kx + (k + 3) = 0$, where $k$ is a constant, has different real roots - Edexcel - A-Level Maths Pure - Question 9 - 2007 - Paper 1

Step 1

Show that $k^2 - 4k - 12 > 0$

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Answer

To show that $k^2 - 4k - 12 > 0$ , we can use the discriminant condition for the quadratic equation to have different real roots. The discriminant is given by:

$D = b^2 - 4ac$

For our equation, comparing with $ax^2 + bx + c = 0$ , we have:

$a = 1$
$b = k$
$c = k + 3$

Thus, substituting into the discriminant formula:

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

This simplifies to:

$D = k^2 - 4k - 12$

For the quadratic to have different real roots, we require:

$D > 0$

Therefore, we need to show:

$k^2 - 4k - 12 > 0$

Step 2

Find the set of possible values of $k$

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Answer

To solve the inequality $k^2 - 4k - 12 > 0$ , we need to find the roots of the equation $k^2 - 4k - 12 = 0$ using the quadratic formula:

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

Here, $a = 1$ , $b = -4$ , and $c = -12$ . Substituting these values gives:

$k = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1}$

Calculating the discriminant:

$D = 16 + 48 = 64$

Now substituting this back into the formula:

$k = \frac{4 \pm 8}{2}$

This results in two roots:

$k = \frac{12}{2} = 6$
$k = \frac{-4}{2} = -2$

To find the intervals where the inequality $k^2 - 4k - 12 > 0$ holds true, we test the intervals:

For $k < -2$ (e.g., $k = -3$ )
For $-2 < k < 6$ (e.g., $k = 0$ )
For $k > 6$ (e.g., $k = 7$ )

By testing these intervals:

In $(-\infty, -2)$ , the expression $k^2 - 4k - 12 > 0$ is true.
In $(-2, 6)$ , the expression is false.
In $(6, \infty)$ , the expression is true.

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

$k < -2 \quad \text{or} \quad k > 6$

The equation $x^2 + kx + (k + 3) = 0$, where $k$ is a constant, has different real roots - Edexcel - A-Level Maths Pure - Question 9 - 2007 - Paper 1

Worked Solution & Example Answer:The equation $x^2 + kx + (k + 3) = 0$, where $k$ is a constant, has different real roots - Edexcel - A-Level Maths Pure - Question 9 - 2007 - Paper 1

Show that $k^2 - 4k - 12 > 0$

Find the set of possible values of $k$

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