A-Level Edexcel Maths Pure Arithmetic Sequences & Series: Find the set of values of $x$ fo

Find the set of values of $x$ for which (a) $2(3x + 4) > 1 - x$ (b) $3x^2 + 8x - 3 < 0$ - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 1

Question 7

Find the set of values of $x$ for which (a) $2(3x + 4) > 1 - x$ (b) $3x^2 + 8x - 3 < 0$

Worked Solution & Example Answer:Find the set of values of $x$ for which (a) $2(3x + 4) > 1 - x$ (b) $3x^2 + 8x - 3 < 0$ - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 1

Step 1

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Answer

Expand the Inequality:

Start with the given inequality: $2(3x + 4) > 1 - x$ Expanding this gives:

$6x + 8 > 1 - x$
Rearrange the Terms:

Collect all $x$ terms on one side and the constant terms on the other:

$6x + x > 1 - 8$ $7x > -7$
Solve for $x$ :

Divide both sides by 7:

$x > -1$

Step 2

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Answer

Form the Quadratic Equation:

Start from the inequality: $3x^2 + 8x - 3 < 0$

To find the critical points, set the quadratic equal to 0:

$(3x + 3)(x - 1) = 0$ This gives:

$x = -1$ and $x = \frac{1}{3}$
Determine the Intervals:

The critical points divide the number line into intervals:
- $(-\infty, -1)$
- $(-1, \frac{1}{3})$
- $(\frac{1}{3}, \infty)$
Test the Intervals:

Choose test values from each interval and evaluate the inequality:
- For $x = -2$ , $$3(-2)^2 + 8(-2) - 3 = 12 - 16 - 3 = -7 < 0$ (true)
- For $x = 0$ , $$3(0)^2 + 8(0) - 3 = -3 < 0$ (true)
- For $x = 1$ , $$3(1)^2 + 8(1) - 3 = 3 + 8 - 3 = 8 > 0$ (false)
Final Solution: The solution to $3x^2 + 8x - 3 < 0$ is:

$-1 < x < \frac{1}{3}$ .