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 $ ext{Solve } \\sqrt{2(1-x)} = x - 1$ Solve for x and y simultaneously: $3^{xy} = 27$ and $x^2 + y^2 = 17$ Determine, without the use of a calculator, the value of: $\frac{1}{\sqrt{1 + \sqrt{2}}} + \frac{1}{\sqrt{2 + \sqrt{3}}} + \frac{1}{\sqrt{3 + \sqrt{4}}} + .. - NSC Mathematics - Question 1 - 2023 - Paper 1

We start with the equation:

$x(3x + 5) = 1$

Rearranging gives:

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

Using the quadratic formula:

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

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

1. Calculate the discriminant: $D = b^2 - 4ac = 5^2 - 4 \times 3 \times (-1) = 25 + 12 = 37$

2. Substitute into the formula:

$x = \frac{{-5 \pm \sqrt{37}}}{6}$

Calculating the two possible values:

1. $x = \frac{{-5 + \sqrt{37}}}{6} \approx 0.18$

2. $x = \frac{{-5 - \sqrt{37}}}{6} \approx -1.85$

Thus, the rounded solutions to TWO decimal places are $x \approx 0.18$ and $x \approx -1.85$.

First, we rearrange the inequality:

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

Factoring gives:

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

To find the critical values, we set the factors equal to zero:

$x - 5 = 0 \implies x = 5$ $x + 3 = 0 \implies x = -3$

Now, we test intervals around the critical values. The intervals are:

1. $(-\infty, -3)$
2. $(-3, 5)$
3. $(5, \infty)$

Testing a value from each interval:

• For $x = -4$: $(-)(-) > 0$ (false)
• For $x = 0$: $(+)(-) < 0$ (true)
• For $x = 6$: $(+)(+) > 0$ (false)

The solution to the inequality is: $-3 < x < 5$

First, square both sides to eliminate the square root:

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

Expanding both sides:

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

Rearranging leads to:

$0 = x^2 - 1 \implies (x - 1)(x + 1) = 0$

Therefore, the critical points are:

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

To verify which solutions satisfy the original equation, we substitute:

• For $x = 1$: LHS $= \sqrt{2(1 - 1)} = 0$ and RHS $= 1 - 1 = 0$ (true)
• For $x = -1$: LHS $= \sqrt{2(1 - (-1))} = \sqrt{4} = 2$ and RHS $= -1 - 1 = -2$ (false)

Thus, the only valid solution is $x = 1$.

Starting with the first equation:

$3^{xy} = 27 \implies 3^{xy} = 3^3 \implies xy = 3 \quad (1)$

Now we substitute $y = \frac{3}{x}$ into the second equation:

$x^2 + \left(\frac{3}{x}\right)^2 = 17$

Multiplying through by $x^2$ gives:

$x^4 - 17x^2 + 9 = 0$

Letting $z = x^2$, we can factor or use the quadratic formula:

To find the roots:

• Using the quadratic formula, we obtain two possible values for $z$ and hence $x$, leading to values for $y$. After computations, we will find $x = 4, y = 1$ or $x = -1, y = 4$.

To simplify:

1. Start with $a_n = \frac{1}{\sqrt{n + \sqrt{n + 1}}}$
2. Rationalize by multiplying the numerator and the denominator by the conjugate: $a_n = \frac{\sqrt{n + 1}}{\sqrt{(n+1) - n}} = \sqrt{n + 1}$

This means we are summing straightforwardly, resulting in a pattern revealing a value of 9 for the entire sum over the specified range of $n$.