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

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

To solve the inequality, first expand the left side:

$2(3x + 4) = 6x + 8$

Now rewrite the inequality:

$6x + 8 > 1 - x$

Next, collect the terms with x on one side and the constant on the other:

$6x + x > 1 - 8$ $7x > -7$

Now, divide both sides by 7:

$x > -1$

Thus, the set of values for which the inequality holds is:

$x > -1$

Step 2

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Answer

To solve this inequality, we first find the critical points by solving the equation:

$3x^2 + 8x - 3 = 0$

We can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, a = 3, b = 8, and c = -3:

$x = \frac{-8 \pm \sqrt{8^2 - 4(3)(-3)}}{2(3)}$ $x = \frac{-8 \pm \sqrt{64 + 36}}{6}$ $x = \frac{-8 \pm \sqrt{100}}{6}$ $x = \frac{-8 \pm 10}{6}$

This gives us two solutions:

$x = \frac{2}{6} = \frac{1}{3}$ and $x = \frac{-18}{6} = -3$

Thus, the critical points are x = -3 and x = \frac{1}{3}.

Next, we analyze the sign of the quadratic in the intervals: (-∞, -3), (-3, \frac{1}{3}), and (\frac{1}{3}, ∞).

Choosing test points, we find:

Thus, the solution set for the inequality is:

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