Los op vir x: 1.1 2x(x + 1) − 7(x + 1) = 0 1.2 x² − 5x − 1 = 0, korrek tot twee desimale plekke - NSC Mathematics - Question 1 - 2017 - Paper 1

Using the quadratic formula:

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

Substituting values: $a = 1$, $b = -5$, $c = -1$:

$$x = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(-1)}}{2(1)} = \frac{5 \pm \sqrt{25 + 4}}{2} = \frac{5 \pm \sqrt{29}}{2}$$

Calculating gives:

1. $x = 5.19$ (to two decimal places)
2. $x = -0.19$ (to two decimal places)

Rearranging gives:

$$4x^2 - 5x + 1 ≥ 0$$

Finding the discriminant:

$$D = b^2 - 4ac = (-5)^2 - 4(4)(1) = 25 - 16 = 9$$

Since D > 0, there are two distinct roots. Setting up:

$$x = \frac{5 \pm \sqrt{9}}{2(4)}$$

Thus, the critical points are:

1. $x_1 = 1$ and $x_2 = \frac{1}{4}$.

Testing intervals shows:

$$x ≤ \frac{1}{4} \text{ or } x ≥ 1$$

First simplify the equation:

$$5(64) - 2^{x1} = 50000$$

Thus,

$$320 - 2^{x1} = 50000$$

Rearranging gives:

$$2^{x1} = 320 - 50000 \Rightarrow 2^{x1} = -49680$$

Since $2^{x1}$ can't be negative, it shows no real solutions exist.

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

$$(2y)^2 + 2(2y) - y - y^2 = 36$$

Expanding gives:

$$4y^2 + 4y - y - y^2 = 36\Rightarrow 3y^2 + 3y - 36 = 0$$

Solving using quadratic formula,

$$$$

This gives:

1. $y = 6$ or $y = -8$

Then substituting back:

1. $x = 12$ or $x = -16$.

For the quadratic equation:

$$ax^2 + bx + c = 0$$

The roots are given by:

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

In this case:

• $a = 1$
• $b = -k$
• $c = k - 1$ The discriminant is:

$$$$

Since $(k - 2)^2 ext{ is non-negative for all } k$, the roots are always real. The roots are rational as they can be expressed in terms of integers.