1.1 Solve for x: 1.1.1 $x^2 - 7x + 12 = 0$ 1.1.2 $x(3x + 5) = 1$ (correct to TWO decimal places) 1.1.3 $x^2 < -2x + 15$ 1.1.4 $\sqrt{2(1-x)} = x - 1$ 1.2 Solve for x and y simultaneously: $3^y = 27$ and $x^2 + y^2 = 17$ 1.3 Determine, without the use of a calculator, the value of: $rac{1}{\sqrt{1 + \sqrt{2}}} + \frac{1}{\sqrt{2 + \sqrt{3}}} + \frac{1}{\sqrt{3 + \sqrt{4}}} + .. - NSC Mathematics - Question 1 - 2023 - Paper 1

Start by rewriting the equation:

$x(3x + 5) = 1$

This can be rearranged to:

$3x^2 + 5x - 1 = 0$

Using the quadratic formula, where $a = 3$, $b = 5$, and $c = -1$:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} = \frac{-5 \pm \sqrt{25 + 12}}{6} = \frac{-5 \pm \sqrt{37}}{6}$

Calculating the two possible values gives approximately:

$x \approx 0.18 \text{ and } x \approx -1.85$

Rearranging the inequality gives:

$x^2 + 2x - 15 < 0$

Factoring:

$(x - 3)(x + 5) < 0$

The critical values are $x = 3$ and $x = -5$. Testing intervals:

1. For $x < -5$: negative
2. For $-5 < x < 3$: positive
3. For $x > 3$: negative

Thus, the solution for the inequality is:

$-5 < x < 3$

To solve this equation, square both sides:

$2(1-x) = (x-1)^2$

Expanding gives:

$2 - 2x = x^2 - 2x + 1$

Rearranging gives:

$x^2 - 1 = 0$

Factoring leads to:

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

Thus, the solutions are:

$x = 1 \text{ and } x = -1$

Since $\sqrt{2(1 - x)}$ must always be non-negative, we only accept $x = 1$.

From the equation $3^y = 27$, we can rewrite $27$ as $3^3$, leading to:

$y = 3$

Next, substituting $y = 3$ into $x^2 + y^2 = 17$ gives:

$x^2 + 3^2 = 17$

Simplifying leads to:

$x^2 + 9 = 17 \Rightarrow x^2 = 8 \Rightarrow x = \pm 2\sqrt{2}$

Thus, we have two sets of solutions:
$(x, y) = (2\sqrt{2}, 3) \text{ and } (-2\sqrt{2}, 3)$.

To evaluate the sum, we observe the general term:

$\frac{1}{\sqrt{n + \sqrt{n+1}}}$

Rationalizing gives:

$\frac{\sqrt{n+1} - \sqrt{n}}{1}$

The successive terms will telescope:

For $n = 1$ to $n = 99$, this becomes:

$(\sqrt{2} - \sqrt{1}) + (\sqrt{3} - \sqrt{2}) + ... + (\sqrt{100} - \sqrt{99})$

The sum simplifies to:

$\sqrt{100} - \sqrt{1} = 10 - 1 = 9$