A-Level Edexcel Maths Pure Differentiation: The equation $20x^2 = 4kx

The equation $20x^2 = 4kx - 13k^2 + 2$, where $k$ is a constant, has no real roots - Edexcel - A-Level Maths Pure - Question 9 - 2018 - Paper 1

Question 9

The equation $20x^2 = 4kx - 13k^2 + 2$, where $k$ is a constant, has no real roots. (a) Show that $k$ satisfies the inequality $$2k^2 + 13k + 20 < 0$$ (b) Find th... show full transcript

Worked Solution & Example Answer:The equation $20x^2 = 4kx - 13k^2 + 2$, where $k$ is a constant, has no real roots - Edexcel - A-Level Maths Pure - Question 9 - 2018 - Paper 1

Step 1

(a) Show that k satisfies the inequality

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Answer

To prove that $k$ satisfies the inequality $2k^2 + 13k + 20 < 0$ , we need to analyze the quadratic expression:

Identify the coefficients: For the quadratic equation in standard form $ax^2 + bx + c$ , here we have:
- $a = 2$ , $b = 13$ , $c = 20$ .
Determine the discriminant: The condition for the quadratic to have no real roots is that the discriminant must be less than zero. The discriminant $riangle$ is given by: $riangle = b^2 - 4ac.$ Substituting the values, we get:

$riangle = 13^2 - 4(2)(20) = 169 - 160 = 9 > 0.$ This indicates that normally it has real roots; however, we must ensure the expression itself is less than zero for certain values.
Verify the inequality: We will check values of $k$ by evaluating the vertex of the quadratic. The vertex occurs at: $k = - rac{b}{2a} = - rac{13}{2(2)} = - rac{13}{4}.$ The maximum value of the quadratic happens here since $a > 0$ . Calculating: $2igg(- rac{13}{4}igg)^2 + 13igg(- rac{13}{4}igg) + 20.$ Evaluating term by term results in a negative or zero, confirming there are intervals where $2k^2 + 13k + 20 < 0.$
Constraints on k: Thus, we conclude that for certain $k$ values derived from $k = - rac{5}{2} ext{ to } k < -10$ , satisfies the inequality derived above.

Step 2

(b) Find the set of possible values for k

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Answer

To find the possible values for $k$ , we need to solve the inequality obtained from part (a):

Set up the inequality: From part (a) we derived: $2k^2 + 13k + 20 < 0$
Finding critical points: Determine when the quadratic equals zero: $2k^2 + 13k + 20 = 0$ We can use the quadratic formula: $k = rac{-b ext{±} ext{√}(b^2 - 4ac)}{2a}$ Plugging in: $k = rac{-13 ext{\pm} ext{\sqrt}(169 - 160)}{4} = rac{-13 ext{\pm} 3}{4}$ This gives us:
- $k_1 = rac{-10}{4} = - rac{5}{2}$
- $k_2 = rac{-16}{4} = -4$ .
Test intervals: The roots divide the $k$ -axis into intervals. We'll test intervals:
- For $k < -4$ : Choose $k = -5$ , it evaluates to positive.
- For $- rac{5}{2} < k < -4$ : Choose $k = -3$ , it evaluates to negative.
- For $k > - rac{5}{2}$ : Choose $k = 0$ , it evaluates to positive.
Determine the valid range: From testing, we need $- rac{5}{2} < k < -4$ to satisfy $2k^2 + 13k + 20 < 0$ . Thus, the set of possible values for $k$ is: $k ext{ in } ig(-5, -4ig)$

The equation $20x^2 = 4kx - 13k^2 + 2$, where $k$ is a constant, has no real roots - Edexcel - A-Level Maths Pure - Question 9 - 2018 - Paper 1

Worked Solution & Example Answer:The equation $20x^2 = 4kx - 13k^2 + 2$, where $k$ is a constant, has no real roots - Edexcel - A-Level Maths Pure - Question 9 - 2018 - Paper 1

(a) Show that k satisfies the inequality

(b) Find the set of possible values for k

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