Given that $x^2 : (3x + 5) = 1 : 2$ find the possible values of x. - Edexcel - GCSE Maths - Question 18 - 2019 - Paper 1

Rearranging the equation gives:

$2x^2 - 3x - 5 = 0$

This is a quadratic equation in standard form.

To find the possible values of x, we use the quadratic formula:

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

where a = 2, b = -3, and c = -5. Plugging in these values gives:

1. Calculate the discriminant:

b$^2 - 4ac = (-3)^2 - 4(2)(-5) = 9 + 40 = 49$

2. Substitute into the formula:

   $x = \frac{3 \pm \sqrt{49}}{2 \times 2}$

   $x = \frac{3 \pm 7}{4}$

Thus, we have:

• $x = \frac{10}{4} = 2.5$
• $x = \frac{-4}{4} = -1$