A-Level Edexcel Maths Pure Equation of a Straight Line: The equation $(k + 3)x^2 + 6x

The equation $(k + 3)x^2 + 6x + k = 5$, where $k$ is a constant, has two distinct real solutions for $x$ - Edexcel - A-Level Maths Pure - Question 11 - 2013 - Paper 3

Question 11

The equation $(k + 3)x^2 + 6x + k = 5$, where $k$ is a constant, has two distinct real solutions for $x$. (a) Show that $k$ satisfies $k^2 - 2k - 24 < 0$ ... show full transcript

Worked Solution & Example Answer:The equation $(k + 3)x^2 + 6x + k = 5$, where $k$ is a constant, has two distinct real solutions for $x$ - Edexcel - A-Level Maths Pure - Question 11 - 2013 - Paper 3

Step 1

Show that k satisfies $k^2 - 2k - 24 < 0$

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Answer

To determine the conditions under which the quadratic equation has two distinct solutions, we must look at the discriminant of the quadratic in the form $ax^2 + bx + c = 0$ . For the equation to have two distinct real solutions, the discriminant must be greater than zero.

Determine coefficients:
In our case, the coefficients are given by the expression:
- $a = (k + 3)$
- $b = 6$
- $c = k - 5$
Discriminant condition:
The discriminant $riangle$ can be calculated as follows:
$riangle = b^2 - 4ac$
Substituting the values gives us:
$riangle = 6^2 - 4(k + 3)(k - 5)$
Simplifying this:
$riangle = 36 - 4(k^2 - 2k - 15)$
$riangle = 36 - 4k^2 + 8k + 60$
$riangle = -4k^2 + 8k + 96$
Setting discriminant > 0:
For two distinct solutions:
$-4k^2 + 8k + 96 > 0$
Dividing by -4 (reversing the inequality):
$k^2 - 2k - 24 < 0$
Thus, we've shown that $k$ must satisfy this condition.

Step 2

Hence find the set of possible values of k

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Answer

Now, let's solve the inequality $k^2 - 2k - 24 < 0$ .

Finding critical points:
We start by solving the associated equation:
$k^2 - 2k - 24 = 0$
Using the quadratic formula:
$k = rac{-b ext{ } ext{±} ext{ } ext{√}(b^2 - 4ac)}{2a}$
where $a = 1$ and $b = -2$ .
Thus,
$k = rac{2 ext{ ± } ext{√}((-2)^2 - 4(1)(-24))}{2(1)} = rac{2 ext{ ± } ext{√}(4 + 96)}{2}$
$k = rac{2 ext{ \pm } ext{\sqrt}(100)}{2} = rac{2 ext{ \pm } 10}{2}$
This gives us the critical points:
- $k = 6$
- $k = -4$
Analyzing the intervals:
We will test the intervals determined by the critical points $(- ext{∞}, -4)$ , $(-4, 6)$ , and $(6, ext{∞})$ :
- For $k < -4$ , say $k = -5$ :
  $(-5)^2 - 2(-5) - 24 = 25 + 10 - 24 = 11 ext{ (not valid)}$
- For $-4 < k < 6$ , say $k = 0$ :
  $0^2 - 2(0) - 24 = -24 < 0 ext{ (valid)}$
- For $k > 6$ , say $k = 7$ :
  $7^2 - 2(7) - 24 = 49 - 14 - 24 = 11 ext{ (not valid)}$
Conclusion:
Therefore, the set of possible values for $k$ that satisfy the inequality is:
$-4 < k < 6$

The equation $(k + 3)x^2 + 6x + k = 5$, where $k$ is a constant, has two distinct real solutions for $x$ - Edexcel - A-Level Maths Pure - Question 11 - 2013 - Paper 3

Worked Solution & Example Answer:The equation $(k + 3)x^2 + 6x + k = 5$, where $k$ is a constant, has two distinct real solutions for $x$ - Edexcel - A-Level Maths Pure - Question 11 - 2013 - Paper 3

Show that k satisfies $k^2 - 2k - 24 < 0$

Hence find the set of possible values of k

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