Solve for x: 1.1.1 $x^2 - 6x = 0$ 1.1.2 $x^2 + 10x + 8 = 0$ (correct to TWO decimal places) 1.1.3 $(1 - x)(x + 2) < 0$ 1.1.4 $rac{ ext{√}x + 18}{x - 2}$ Solve simultaneously for x and y: $x + y = 3$ and $2x^2 + 4xy - y = 15$ If n is the largest integer for which $n^{200} < 530^0$, determine the value of n. - NSC Mathematics - Question 1 - 2020 - Paper 1

For this quadratic equation, we can use the quadratic formula:

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

Here, $a = 1$, $b = 10$, and $c = 8$. Substituting these values in gives:

$x = \frac{-10 \pm \sqrt{10^2 - 4(1)(8)}}{2(1)} = \frac{-10 \pm \sqrt{100 - 32}}{2} = \frac{-10 \pm \sqrt{68}}{2}$

Calculating further:

$x = \frac{-10 \pm 8.246}{2}$

Thus, the two solutions are:

1. $x \approx \frac{-10 + 8.246}{2} \approx -0.88$ (rounded to two decimal places)
2. $x \approx \frac{-10 - 8.246}{2} \approx -9.12$ (rounded to two decimal places).

To solve the inequality $(1 - x)(x + 2) < 0$, we first find the critical values by setting each factor to zero:

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

Next, we test intervals around these critical points:

• For $x < -2$, both factors are positive, so the product is positive.
• For $-2 < x < 1$, $(1 - x)$ is positive and $(x + 2)$ is negative, so the product is negative.
• For $x > 1$, both factors are negative, so the product is positive.

Thus, the solution is:

$-2 < x < 1$

To solve $ext{√}x + 18 = x - 2$, we first square both sides to eliminate the square root:

$\text{√}x + 18 = x - 2$

Rearranging gives:

$\text{√}x = x - 2 - 18$

Squaring both sides results in:

$x + 18x - 36 = 0$

This can be simplified to:

$x^2 - 18x + 36 = 0$

Finding solutions involves factoring or using the quadratic formula. The valid solutions are obtained by ensuring the restrictions from both sides are met.

From the first equation, we have:

$y = 3 - x$

Substituting this into the second equation:

$2x^2 + 4x(3 - x) - (3 - x) = 15$

Expanding this gives:

$2x^2 + 12x - 4x^2 - 3 + x - 15 = 0$

Rearranging results in:

$-2x^2 + 13x - 18 = 0$

Dividing through by -1:

$2x^2 - 13x + 18 = 0$

Finding solutions via the quadratic formula then gives two solutions for $x$ and corresponding values of $y$.

Since $530^0 = 1$, we have:

$n^{200} < 1$

The solution implies:

$n < 1^{1/200} = 1$

Thus, the largest integer $n$ that satisfies this equation is:

$n = 0$