9 (a) Expand and simplify $(x - 2)(2x + 3)(x + 1)$ (b) Find the value of $n$ - Edexcel - GCSE Maths - Question 10 - 2018 - Paper 3

To find the value of $n$, we equate the exponents in the expression:

$\frac{y^3 \cdot y^n}{y^2} = y^3$

This implies that:

$3 + n - 2 = 3$

Solving for $n$ gives:

$n = 2$.

To solve the quadratic equation $5x^2 - 4x - 3 = 0$, we can use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Where $a = 5$, $b = -4$, and $c = -3$.

1. Calculate the discriminant:
   $b^2 - 4ac = (-4)^2 - 4(5)(-3) = 16 + 60 = 76$.

2. Now, substituting into the quadratic formula:
   $x = \frac{4 \pm \sqrt{76}}{10}$

3. Simplifying further, we find:
   $x = \frac{4 \pm 2\sqrt{19}}{10} = \frac{2 \pm \sqrt{19}}{5}$.

4. The approximate solutions of $x$ are:
   $x \approx 1.27$ and $x \approx -0.47$, rounded to 3 significant figures.