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The equation $2x^2 - 3x - (k + 1) = 0$, where $k$ is a constant, has no real roots - Edexcel - A-Level Maths Pure - Question 7 - 2007 - Paper 2

Question 7

The equation $2x^2 - 3x - (k + 1) = 0$, where $k$ is a constant, has no real roots. Find the set of possible values of $k$.

Worked Solution & Example Answer:The equation $2x^2 - 3x - (k + 1) = 0$, where $k$ is a constant, has no real roots - Edexcel - A-Level Maths Pure - Question 7 - 2007 - Paper 2

Step 1

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Answer

For the quadratic equation $ax^2 + bx + c = 0$ , the condition for no real roots is given by the discriminant, defined as $D = b^2 - 4ac$ .

Here, the coefficients are:

Thus, we express the discriminant as:

$D = (-3)^2 - 4(2)(-(k + 1))$ $D = 9 + 8(k + 1)$ $D = 9 + 8k + 8$ $D = 8k + 17.$

To ensure there are no real roots, we require:

ightarrow 8k + 17 < 0$$

Step 2

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Answer

Now, we solve the inequality:

$8k + 17 < 0$

Subtracting 17 from both sides yields:

$8k < -17$

Dividing by 8 gives:

$k < -\frac{17}{8}$

This simplifies to:

$k < -2.125.$

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

$k \in (-\infty, -2.125).$

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