1.1 Solve for x: 1.1.1 3x² + 5x = 0 1.1.2 4x² + 3x - 5 = 0 (answers correct to TWO decimal places) 1.1.3 (x - 1)² - 9 ≥ 0 1.1.4 5x² - 5 = 0 1.1.5 \( \frac{x}{\sqrt{20 - x}} = 1 \) 1.2 Solve for x and y simultaneously: x + y = 9 and 2x² - y² = 7 1.3 Given: P = (1 - a)(1 + a²)(1 + a¹²) - NSC Mathematics - Question 1 - 2024 - Paper 1

Using the quadratic formula, where ( a = 4, b = 3, c = -5 ):

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

Plugging in the values gives:

$x = \frac{{-3 \pm \sqrt{{3^2 - 4 \cdot 4 \cdot (-5)}}}}{{2 \cdot 4}}$

Calculating the discriminant:

$= \sqrt{{9 + 80}} = \sqrt{89}$

Thus:

1. ( x = \frac{{-3 + \sqrt{89}}}{8} \approx 0.80 )
2. ( x = \frac{{-3 - \sqrt{89}}}{8} \approx -1.55 )

Rearranging gives:

$(x - 1)² ≥ 9$

Taking square roots:

$|x - 1| ≥ 3$

Thus, we have two cases:

1. ( x - 1 ≥ 3 \Rightarrow x ≥ 4 )
2. ( x - 1 ≤ -3 \Rightarrow x ≤ -2 )

Final solutions are: ( x ≤ -2 ) or ( x ≥ 4 )

We can factor this equation:

$5(x² - 1) = 0$

This means:

1. ( x = 0 )
2. ( x² = 1 \Rightarrow x = 1 \text{ or } x = -1 )

To solve this, square both sides:

$x² = 20 - x$

Rearranging gives:

$x² + x - 20 = 0$

Factoring yields:

$(x + 5)(x - 4) = 0$

Thus, ( x = -5 ) or ( x = 4 )

From the first equation, express y:

$y = 9 - x$

Substituting into the second equation:

$2x² - (9 - x)² = 7$

Expanding gives:

$2x² - (81 - 18x + x²) = 7$

Combining like terms results in:

$x² + 18x - 88 = 0$

Using the quadratic formula results in potential x-values: ( x = 5 ) or ( x = -22 ). Substituting back gives corresponding y-values: ( y = 4 ) or ( y = 31 ).

Calculating P:
( P = (1 - a)(1 + a²)(1 + a¹²) )
Expand P to find: ( P = (1 - a)(1 + a² + a¹² + a^{14}) )
Next, calculate T: ( T = (1 + a)(1 + a^{12}) )
Combining, we find: ( P \times T = (1 - a)(1 + a)(1 + a²)(1 + a^{12}) )