A-Level Edexcel Maths Pure Applications of Differentiation: 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 9 - 2008 - Paper 2

Question 9

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 9 - 2008 - Paper 2

Step 1

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

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Answer

To show that the equation has no real solutions, we first rearrange it:

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

Next, we apply the discriminant condition for quadratic equations, which states that for the quadratic $ax^2 + bx + c = 0$ to have no real solutions, the discriminant must be negative:

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

Here, $a = 1$ , $b = k$ , and $c = 8 - k$ . Thus, the discriminant becomes:

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

Calculating this gives:

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

This confirms that $k$ must satisfy the inequality $k^2 + 4k - 32 < 0$ .

Step 2

Hence find the set of possible values of $k$

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Answer

To find the set of possible values of $k$ , we solve the quadratic inequality:

First, we find the roots of the equation $k^2 + 4k - 32 = 0$ using the quadratic formula:

$k = rac{-b \\pm \\sqrt{b^2 - 4ac}}{2a} = rac{-4 \\pm \\sqrt{16 + 128}}{2} = rac{-4 \\pm 12}{2}$

Calculating the roots yields:

$k = rac{8}{2} = 4$
$k = rac{-16}{2} = -8$

The roots are $k = -8$ and $k = 4$ .

To determine where the quadratic is less than zero, we test intervals. The intervals are $(-\\infty, -8)$ , $(-8, 4)$ , and $(4, +\\infty)$ .

Choosing test points in each interval:

For $k = -9$ in $(-\\infty, -8)$ , we have $(-9)^2 + 4(-9) - 32 = 81 - 36 - 32 = 13 > 0$ .
For $k = 0$ in $(-8, 4)$ , we have $0^2 + 4(0) - 32 = -32 < 0$ .
For $k = 5$ in $(4, +\\infty)$ , we have $5^2 + 4(5) - 32 = 25 + 20 - 32 = 13 > 0$ .

Therefore, the quadratic is less than zero in the interval $(-8, 4)$ :

The set of possible values of $k$ is

$-8 < k < 4$

The equation $$x^2 + kx + 8 = k$$ has no real solutions for $x$ - Edexcel - A-Level Maths Pure - Question 9 - 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 9 - 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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