A-Level Edexcel Maths Pure Polynomials: The equation $$x^2 + kx + 8 = k

The equation $$x^2 + kx + 8 = k$$ has no real solutions for $x$ - Edexcel - A-Level Maths Pure - Question 10 - 2008 - Paper 2

Question 10

The equation $$x^2 + kx + 8 = k$$ has no real solutions for $x$. (a) Show that $k$ satisfies $k^2 + 4k - 32 < 0$. (b) Hence find the set of possible values of ... show full transcript

Worked Solution & Example Answer:The equation $$x^2 + kx + 8 = k$$ has no real solutions for $x$ - Edexcel - A-Level Maths Pure - Question 10 - 2008 - Paper 2

Step 1

Show that $k$ satisfies $k^2 + 4k - 32 < 0$

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Answer

To show that the inequality holds, we can start from the equation:

$x^2 + kx + 8 - k = 0$

This needs to have no real solutions, which is determined by the discriminant being less than zero:

$D = b^2 - 4ac < 0$

In this case, we have:

$a = 1$ ,
$b = k$ ,
$c = 8 - k$ .

Thus, we compute the discriminant:

$D = k^2 - 4(1)(8 - k) = k^2 - 32 + 4k$

Now, we want to ensure:

$k^2 + 4k - 32 < 0$

This indicates that the expression forms a quadratic whose solutions will tell us the range of $k$ for which this inequality is valid.

Step 2

Hence find the set of possible values of $k$

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Answer

To find the possible values of $k$ , we first determine the roots of the quadratic equation:

$k^2 + 4k - 32 = 0$

Using the quadratic formula:

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

Here, substituting $a = 1$ , $b = 4$ , and $c = -32$ gives:

$k = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-32)}}{2 \cdot 1} = \frac{-4 \pm \sqrt{16 + 128}}{2} = \frac{-4 \pm \sqrt{144}}{2}$

This simplifies to:

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

Calculating the solutions:

First root: $k = \frac{8}{2} = 4$
Second root: $k = \frac{-16}{2} = -8$

Thus, the range for the inequality $k^2 + 4k - 32 < 0$ lies between the roots:

$-8 < k < 4$

So the set of possible values for $k$ is:

$k \in (-8, 4)$ .

The equation $$x^2 + kx + 8 = k$$ has no real solutions for $x$ - Edexcel - A-Level Maths Pure - Question 10 - 2008 - Paper 2

Worked Solution & Example Answer:The equation $$x^2 + kx + 8 = k$$ has no real solutions for $x$ - Edexcel - A-Level Maths Pure - Question 10 - 2008 - Paper 2

Show that $k$ satisfies $k^2 + 4k - 32 < 0$

Hence find the set of possible values of $k$

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