Solve for x: 1.1.1 $x^{2}-x-12=0$ 1.1.2 $x(x+3)-1=0$ (Leave your answer in simplest surd form.) 1.1.3 $x(4-x)<0$ 1.1.4 $x= \frac{a^{2}+a-2}{a-1}$ if $a=888888888$ - NSC Mathematics - Question 1 - 2016 - Paper 1

To solve the equation, we first rearrange it:

$x(x+3)=1$

Next, expand and rearrange:

$x^{2}+3x-1=0$

Using the quadratic formula where: $a=1, b=3, c=-1$

We have:

$x = \frac{-b\pm\sqrt{b^{2}-4ac}}{2a} = \frac{-3\pm\sqrt{3^{2}-4(1)(-1)}}{2(1)} = \frac{-3\pm\sqrt{9+4}}{2} = \frac{-3\pm\sqrt{13}}{2}$

Thus the solutions in simplest surd form are: $x = \frac{-3+\sqrt{13}}{2} \quad \text{and} \quad x = \frac{-3-\sqrt{13}}{2}$

To solve the inequality, we first identify the critical points by setting:

$x(4-x)=0$

This gives: $x=0 \text{ or } x=4$

Next, we'll determine the intervals to test:

• Interval 1: $(-\infty, 0)$
• Interval 2: $(0, 4)$
• Interval 3: $(4, \infty)$

Testing these intervals, we find:

• For Interval 1, choose $x=-1$: $(-1)(4-(-1))=5>0$
• For Interval 2, choose $x=2$: $(2)(4-2)=4<0$
• For Interval 3, choose $x=5$: $(5)(4-5)=-5<0$

Thus, the solution to the inequality is: $x \in (0, 4)$.

Substitute $a=888888888$ into the equation:

$x= \frac{(888888888)^{2}+888888888-2}{888888888-1}$

Calculating this expression yields:

$x=888888888$.

From the first equation, we can express $y$ in terms of $x$:

$y=2x-7$

Substituting this into the second equation:

$x^{2}-(2x-7)x+3(2x-7)^{2}=15$

Expanding and simplifying gives:

$x^{2}-2x^{2}+7x+3(4x^{2}-28x+49)=15$

Combine like terms:

$11x^{2}-19x-60=0$

Now, we can solve this quadratic equation using the quadratic formula. Solving for $x$ gives:

$x = 3 \text{ or } x= -1$

Now substituting these values back into $y=2x-7$ to find corresponding $y$:

For $x=3$: $y=2(3)-7= -1$

For $x=-1$: $y=2(-1)-7= -9$

Thus, the solutions are $(x, y) = (3, -1)$ and $(-1, -9)$.

To analyze the range of the function, we can first find its derivative:

$y^{'}=1+\frac{1-x}{x^{2}}$

Setting the derivative to zero to find critical points:

$1+\frac{1-x}{x^{2}}=0 \Rightarrow x^{2}=-1+x \Rightarrow x^{2}-x+1=0$

This has no real solutions showing that the function is always increasing for $x>0$ and $x<0$. Now as $x \to 0^{+}$, we find $y \to \infty$ and as $x \to \infty$, $y \to \infty$. Hence the minimum point can be found by testing values around $x=1$:

Evaluating at $x=1$ gives:

$y=1+1=2$

Thus, the range of the function is: $y \geq 2$, hence the range of the function is $[2, \infty)$.