The diagram shows a parallelogram - Edexcel - GCSE Maths - Question 23 - 2019 - Paper 2

To find the range of values of x that satisfy $2x^2 - 21x + 40 < 0$, we first factor the quadratic:

$$2x^2 - 21x + 40 = 0$$

Using the quadratic formula:

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

where:

• $a = 2$, $b = -21$, and $c = 40$,

This results in:

$$x = \frac{21 \pm \sqrt{(-21)^2 - 4(2)(40)}}{2(2)}$$

Calculating the discriminant:

$$(-21)^2 - 4(2)(40) = 441 - 320 = 121$$

Thus, we find:

$$x = \frac{21 \pm 11}{4}$$

Calculating the roots:

1. $$$$

x = \frac{21 + 11}{4} = 8

2. $$ x = \frac{21 - 11}{4} = 2.5

The quadratic will be negative between the roots, thus:

$$2.5 < x < 8.$$

Therefore, the range of possible values of x is:

2.5 < x < 8.$$