Solve for x: 1.1 2x(x + 1) - 7(x + 1) = 0 1.2 x² - 5x - 1 = 0, correct to two decimal places - NSC Mathematics - Question 1 - 2017 - Paper 1

Using the quadratic formula:

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

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

Calculating the discriminant:

$$b^2 - 4ac = (-5)^2 - 4(1)(-1) = 25 + 4 = 29$$

Thus,

$$x = \frac{5 \pm \sqrt{29}}{2}$$

Calculating the values gives approximately $x = 5.19$ and $x = -0.19$.

Rearranging gives:

$$4x² - 5x + 1 ≥ 0$$

Using the quadratic formula to find roots:

$$x = \frac{5 \pm \sqrt{(-5)^2 - 4(4)(1)}}{2(4)} = \frac{5 \pm \sqrt{9}}{8}$$

Thus,

$$x = 1 \text{ or } x = \frac{1}{4}$$

The solution to the inequality is $x \leq \frac{1}{4}$ or $x \geq 1$.

Calculating $5 × 4³$:

$$5 × 64 = 320$$

Substituting:

$$320 - 100 - 2x + 1 = 50000$$

This simplifies to:

$$221 - 2x = 50000\ 2x = 221 - 50000\ x = \frac{49779}{2} = 23889.5$$

We have the equations:

1. $x = 2y$
2. $x² + 2x - y² = 36$

Substituting $x = 2y$ into the second equation:

$$(2y)² + 2(2y) - y² = 36$$

This simplifies to:

$$4y² + 4y - y² = 36\ 3y² + 4y - 36 = 0$$

Using the quadratic formula to solve for $y$:

$$y = \frac{-4 \pm \sqrt{4^2 - 4(3)(-36)}}{2(3)} = \frac{-4 \pm \sqrt{144}}{6}\ y = 6 ext{ or } y = -8$$

Thus:

Replacing back, we find $x = 12$ or $x = -16$.

Using the discriminant for the quadratic:

$$D = b^2 - 4ac = (-k)^2 - 4(1)(k - 1) = k^2 - 4k + 4 = (k - 2)²$$

Since $(k - 2)²$ is always non-negative, the roots are real. For rationality, since $(k - 2)²$ is a perfect square, the roots are also rational for all real values of k.