Given: $f(x) = x^2 - 3x - 10$ Solve for $x$ if: 1.1.1 $f(x) = 0$ 1.1.2 $f(x) < 0$ and represent the solution on a number line - NSC Technical Mathematics - Question 1 - 2022 - Paper 1

The roots found are $x = -2$ and $x = 5$.
To determine where $f(x) < 0$, we test the intervals defined by the roots:

• For $x < -2$: Choose $x = -3$:
  $f(-3) = (-3)^2 - 3(-3) - 10 = 9 + 9 - 10 = 8 > 0$
• For $-2 < x < 5$: Choose $x = 0$:
  $f(0) = 0^2 - 3(0) - 10 = -10 < 0$
• For $x > 5$: Choose $x = 6$:
  $f(6) = 6^2 - 3(6) - 10 = 36 - 18 - 10 = 8 > 0$
  Thus, the solution for $f(x) < 0$ is $-2 < x < 5$.
  On a number line, this is represented as:
  $(-2, 5)$.

Rearranging the equation gives:
$2x^2 + 7x - 11 = 0$
Applying the quadratic formula:
$x = \frac{-7 \pm \sqrt{7^2 - 4(2)(-11)}}{2(2)}$
The discriminant is:
$49 + 88 = 137$
Thus,
$x = \frac{-7 \pm \sqrt{137}}{4}$
Calculating the two solutions gives:

1. $x_1 \approx 0.85$
2. $x_2 \approx -6.35$
   Correct to TWO decimal places: $x \approx 0.85, -6.35$.

From the first equation, solve for $y$:
$y = x - 1$
Substituting into the second equation:
$x - 1 + 7 = x^2 + 2x$
Simplifying gives:
$x - 1 + 7 = x^2 + 2x\implies x^2 + x - 6 = 0$
Factoring yields:
$(x - 2)(x + 3) = 0$
Thus, $x = 2$ or $x = -3$.
If $x = 2$: $y = 2 - 1 = 1$.
If $x = -3$: $y = -3 - 1 = -4$.
Thus, the solutions are $(x, y) = (2, 1)$ and $(-3, -4)$.

Starting from:
$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$
Taking the reciprocal gives:
$R_p = \frac{R_1R_2}{R_1 + R_2}$
Thus, $R_p$ is made the subject of the formula.

Using the formula derived:
$R_p = \frac{R_1R_2}{R_1 + R_2} = \frac{40 \times 45}{40 + 45} = \frac{1800}{85} \approx 21.18 \Omega$
Thus, the total resistance is approximately $21.18 \Omega$.

First, convert both binary numbers to decimal:

• $1101100_2 = 108$
• $1100_2 = 12$
  Adding these gives $108 + 12 = 120$.
  Now convert $120$ back to binary, which yields $1111000_2$.
  Thus, $1101100_2 + 1100_2 = 1111000_2$.