Here is a rectangle and a right-angled triangle - Edexcel - GCSE Maths - Question 23 - 2017 - Paper 1

The triangle has a base of (2x) and a height of (x). Thus, the area of the triangle, (A_t), is:

$A_t = \frac{1}{2} \times 2x \times x = x^2$

To find the values of (x) for which the area of the rectangle is greater than the area of the triangle, we set up the inequality:

$3x^2 - 3x - 2 > x^2$

Rearranging gives:

$2x^2 - 3x - 2 > 0$

We need to solve the quadratic equation (2x^2 - 3x - 2 = 0) using the quadratic formula:

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

Here, (a = 2), (b = -3), and (c = -2):

1. Calculate the discriminant:

   $D = (-3)^2 - 4 \times 2 \times (-2) = 9 + 16 = 25$

2. Solve for (x):

   $x = \frac{3 \pm 5}{4}$

   Which gives us two solutions:

   $x = \frac{8}{4} = 2 \quad \text{and} \quad x = \frac{-2}{4} = -0.5$

The critical points are (x = 2) and (x = -0.5). To determine the intervals where the inequality (2x^2 - 3x - 2 > 0) holds, we can test intervals formed by these critical points.

• For (x < -0.5): test with (x = -1): (2(-1)^2 - 3(-1) - 2 = 2 + 3 - 2 > 0) (True)
• For (-0.5 < x < 2): test with (x = 0): (2(0)^2 - 3(0) - 2 = -2 < 0) (False)
• For (x > 2): test with (x = 3): (2(3)^2 - 3(3) - 2 = 18 - 9 - 2 > 0) (True)

Thus, the solution to the inequality is (x < -0.5) or (x > 2). Since all measurements are in centimeters, we consider (x > 2) to ensure positive dimensions.