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: $\frac{1}{\sqrt{1 + \frac{1}{2}}} + \frac{1}{\sqrt{2 + \frac{1}{3}}} + \frac{1}{\sqrt{3 + \frac{1}{4}}} + .. - NSC Mathematics - Question 1 - 2023 - Paper 1

Rearranging the equation gives us:

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

Using the quadratic formula:

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

where $a = 3, b = 5, c = -1$:

1. Calculate the discriminant:
   • $b^2 - 4ac = 5^2 - 4(3)(-1) = 25 + 12 = 37$
2. Substitute into the quadratic formula:
   • $x = \frac{-5 \pm \sqrt{37}}{6}$
3. This gives two solutions:
   • $x \approx -0.18$
   • $x \approx -1.85$ (to two decimal places)

Rearranging gives:

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

Factoring, we have:

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

The critical values are عند $x = -5$ و $x = 3$. We analyze the intervals:

• For $x < -5$, ($x + 5 < 0$, $x - 3 < 0$) => (both negative)
• For $-5 < x < 3$, ($x + 5 > 0$, $x - 3 < 0$) => (one positive, one negative)
• For $x > 3$, ($x + 5 > 0$, $x - 3 > 0$) => (both positive) Thus, the solution is:

$-5 < x < 3$.

Squaring both sides:

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

Expanding gives:

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

Rearranging leads to:

$$x^2 - 1 = 0$$

This factors as:

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

Thus, the solutions are:

• $x = 1$
• $x = -1$ We also check these solutions in the original equation to verify if they lie within the valid range.

Rewriting $3^y = 27$ gives:

y = 3$$ Substituting $y$ into the second equation:

x^2 + 3^2 = 17$$

x^2 + 9 = 17$$ This simplifies to:

x^2 = 8$$ Thus:

x = \pm \sqrt{8} = \pm 2\sqrt{2}$$ The pairs $(x, y)$ are: - $(2\sqrt{2}, 3)$ - $(-2\sqrt{2}, 3)$.

This expression is a sum of terms of the form:

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

First, simplify each term in the series:

1. Rationalizing gives a general term of: $\frac{1}{\sqrt{1 + \frac{1}{n}}} = \frac{1}{\sqrt{\frac{n + 1}{n}}} = \sqrt{\frac{n}{n + 1}}$
2. The sum converges as: $1 \rightarrow \infty \text{ for } n = 1 ext{ to } 100$
3. Final evaluation leads to the result of: $9$