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: \( P = (1 - a)(1 + a)(1 + a²)(1 + a^{12}) \) Determine the value of \( P \times T \) in terms of a. - NSC Mathematics - Question 1 - 2024 - Paper 1

We apply the quadratic formula:

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

Here, a = 4, b = 3, c = -5.

Calculating the discriminant:

$b^2 - 4ac = 3^2 - 4 \cdot 4 \cdot (-5) = 9 + 80 = 89$

Now substituting into the formula:

$x = \frac{-3 \pm \sqrt{89}}{8}$

Calculating:

1. $x_1 = \frac{-3 + \sqrt{89}}{8} \approx 0.80$
2. $x_2 = \frac{-3 - \sqrt{89}}{8} \approx -1.55$.

To solve this inequality, we start by rewriting it:

$(x - 1)² ≥ 9$

Taking the square root of both sides gives:

$|x - 1| ≥ 3$

This results in two cases:

1. $x - 1 ≥ 3 \Rightarrow x ≥ 4$
2. $x - 1 ≤ -3 \Rightarrow x ≤ -2$

Thus, the solution is: $x ≤ -2 \text{ or } x ≥ 4$.

Factoring the left side results in:

$5(x² - 1) = 0$

This simplifies to:

$5(x - 1)(x + 1) = 0$

Thus, the solutions are:

1. $x - 1 = 0 \Rightarrow x = 1$
2. $x + 1 = 0 \Rightarrow x = -1$.

We start by isolating the surd:

$x = \sqrt{20 - x}$

Squaring both sides leads to:

$x^2 = 20 - x$

Rearranging yields:

$x^2 + x - 20 = 0$

Factoring gives:

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

Thus, the solutions are:

1. $x + 5 = 0 \Rightarrow x = -5$
2. $x - 4 = 0 \Rightarrow x = 4$.

From the first equation, we express y in terms of x:

$y = 9 - x$

Substituting into the second equation:

$2x² - (9 - x)² = 7$

Expanding the equation:

$2x² - (81 - 18x + x²) = 7$

Simplifying leads to:

$x² + 18x - 88 = 0$

Using the quadratic formula gives:

$x = 22 \text{ or } x = -4$

Substituting back to find y:

1. If $x = 22: y = -13$
2. If $x = -4: y = 13$.

To find ( P \times T ), we first assume ( T = (1 + a)(1 + a²)(1 + a^{12}) ).

Thus, ( P \times T = [(1 - a)(1 + a)(1 + a²)(1 + a^{12})] \times [(1 + a)(1 + a²)(1 + a^{12})] )

This expands to:

( P \times T = (1 - a)(1 + a)^2(1 + a²)²(1 + a^{12})² ).

Hence, the expression for ( P \times T ) in terms of a is: ( P \times T = (1 - a)(1 + a)^2(1 + a²)²(1 + a^{12})² ).