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 in the expression $\frac{y^3 \times y^n}{y^2} = y^3$, first simplify the expression on the left-hand side:

$\frac{y^{3+n}}{y^2} = y^{3+n-2} = y^{1+n}.$ Set the left-hand side equal to the right-hand side:

$y^{1+n} = y^3.$ For this equation to hold true, the exponents must be equal:

$1+n = 3,$ thus:

$n = 3 - 1 = 2.$

The value of n is 2.

To solve the equation 5x² − 4x − 3 = 0, we will use the quadratic formula:

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

First, calculate the discriminant:

$b^2 - 4ac = (-4)^2 - 4(5)(-3) = 16 + 60 = 76.$

Now substituting back into the quadratic formula yields:

$x = \frac{-(-4) \pm \sqrt{76}}{2 \cdot 5} = \frac{4 \pm 2\sqrt{19}}{10} = \frac{2 \pm \sqrt{19}}{5}.$

Calculating the two roots:

1. $x_1 = \frac{2 + \sqrt{19}}{5} \approx 1.27$
2. $x_2 = \frac{2 - \sqrt{19}}{5} \approx -0.47.$

Thus, the solutions to the equation are approximately 1.27 and -0.47, correct to 3 significant figures.