1.1 Solve for x, in each of the following: 1.1.1 $2x^2 - 7x = 0$ 1.1.2 $4x + \frac{4}{x} + 11 = 0; \; x \neq 0$ (correct to TWO decimal places) 1.1.3 $(2x - 1)(x - 3) > 0$ 1.1.4 $3^{x} \cdot 3^{1} = 27^{x}$ 1.2 Solve simultaneously for x and y in the following equations: $3 + y = 2x$ and $4x^2 + y^2 = 2xy + 7$ 1.3 Given: $f(x) = x^3 - 4x^2 - 2x + 20 = (x + 2)(x^2 - 6x + 10)$ Prove that $f(x)$ has only one real root. - NSC Mathematics - Question 1 - 2016 - Paper 1

Multiplying through by x to eliminate the fraction results in:

$4x^2 + 11x + 4 = 0$

Using the quadratic formula, $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 4$, $b = 11$, and $c = 4$:

$x = \frac{-11 \pm \sqrt{11^2 - 4 \cdot 4 \cdot 4}}{2 \cdot 4} = \frac{-11 \pm \sqrt{81}}{8} = \frac{-11 \pm 9}{8}$

Calculating these gives:

$x_1 = \frac{-2}{8} = -0.25 \quad \text{and} \quad x_2 = \frac{-20}{8} = -2.5$

Rounded to TWO decimal places, the solutions are:

$x = -0.25 \quad \text{and} \quad x = -2.50$

To solve the inequality, we identify the critical points where the expression equals zero:

$2x - 1 = 0 \quad \Rightarrow \quad x = \frac{1}{2}$ $x - 3 = 0 \quad \Rightarrow \quad x = 3$

Testing intervals between the critical points:

• For $x < \frac{1}{2}$, choose $x = 0$: $(-1)(-3) > 0$ (true)
• For $\frac{1}{2} < x < 3$, choose $x = 1$: $(1)(-2) < 0$ (false)
• For $x > 3$, choose $x = 4$: $(7)(1) > 0$ (true)

Thus, the solution set is:

$x < \frac{1}{2} \quad \text{or} \quad x > 3$

Rewriting $27$ as $3^3$, we have:

$3^{x} \cdot 3^{1} = (3^3)^{x} \implies 3^{x + 1} = 3^{3x}$

Equating the exponents:

$x + 1 = 3x$

Solving for x:

$1 = 2x \quad \Rightarrow \quad x = \frac{1}{2}$

From the first equation, expressing y in terms of x gives:

$y = 2x - 3$

Substituting into the second equation:

$4x^2 + (2x - 3)^2 = 2x(2x - 3) + 7$

Expanding and simplifying leads to:

$4x^2 + (4x^2 - 12x + 9) = (4x^2 - 6x) + 7$

Simplifying further gives:

$8x^2 - 12x + 9 = 4x^2 - 6x + 7$

Bringing everything to one side:

$4x^2 - 6x + 2 = 0$

Using the quadratic formula yields roots:

$x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 4 \cdot 2}}{2 \cdot 4} = \frac{6 \pm 2}{8}$

Giving us:

$x = 1 \quad \text{or} \quad x = \frac{1}{2}$

Substituting these back to find corresponding y-values gives:

For $x = 1$, $y = -1$. Thus $(x, y) = (1, -1)$.

First, we will analyze the quadratic factor:

$x^2 - 6x + 10$

Calculating its discriminant:

$D = b^2 - 4ac = (-6)^2 - 4(1)(10) = 36 - 40 = -4$

Since the discriminant is negative, this quadratic has no real roots. Thus, the only real root of $f(x)$ comes from the linear factor:

$x + 2 = 0 \quad \Rightarrow \quad x = -2$

Therefore, $f(x)$ has exactly one real root.