QUESTION 1 Solve for $$ x $$ : 1.1.1 $$ \quad x^{2}-7 x+12=0 $$ $$ 1.1 .2 \quad x(3 x+5)=1 \quad $$ (correct to TWO decimal pla $$ 1.1 .3 \quad x^{2}

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QUESTION 1 Solve for $$ x $$ : 1.1.1 $$ \quad x^{2}-7 x+12=0 $$ $$ 1.1 .2 \quad x(3 x+5)=1 \quad $$ (correct to TWO decimal pla $$ 1.1 .3 \quad x^{2}<-2 x+15 $$ $$ 1.1 .4 \quad \sqrt{2(1-x)}=x-1 $$ $$ 1.2 \quad $$ Solve for $$ x $$ and $$ y $$ simultaneously: $$ 3^{x+y}=27 \quad $$ and $$ \quad x^{2}+y^{2}=17 $$

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**1.1.1**  
We need to solve  
$$
x^2 - 7x + 12 = 0.
$$  
Factor the quadratic:  
$$
x^2 - 7x + 12 = (x-3)(x-4)=0.
$$  
Thus,  
$$
x=3 \quad \text{or} \quad x=4.
$$

---

**1.1.2**  
We need to solve  
$$
x(3x+5)=1.
$$  
First, expand and rearrange to obtain a quadratic equation:  
$$
3x^2 + 5x - 1 = 0.
$$  
Using the quadratic formula  
$$
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a},
$$  
with $$a=3$$, $$b=5$$, and $$c=-1$$, we have  
$$
x = \frac{-5 \pm \sqrt{5^2-4\cdot 3\cdot(-1)}}{2\cdot 3} = \frac{-5 \pm \sqrt{25+12}}{6} = \frac{-5 \pm \sqrt{37}}{6}.
$$  
Approximating to two decimal places:  
$$
\sqrt{37} \approx 6.08,
$$  
so  
$$
x \approx \frac{-5+6.08}{6} \approx \frac{1.08}{6} \approx 0.18,
$$  
and  
$$
x \approx \frac{-5-6.08}{6} \approx \frac{-11.08}{6} \approx -1.85.
$$

---

**1.1.3**  
We need to solve the inequality  
$$
x^2 < -2x + 15.
$$  
First, bring all terms to one side:  
$$
x^2 + 2x - 15 < 0.
$$  
Solve the corresponding quadratic equation  
$$
x^2 + 2x - 15 = 0.
$$  
The discriminant is  
$$
\Delta = 2^2 - 4\cdot 1\cdot (-15) = 4 + 60 = 64.
$$  
The roots are  
$$
x = \frac{-2 \pm \sqrt{64}}{2} = \frac{-2\pm 8}{2}.
$$  
This gives  
$$
x = 3 \quad \text{and} \quad x = -5.
$$  
Since the quadratic opens upward, the inequality holds between the roots. Thus,  
$$
-5 < x < 3.
$$

---

**1.1.4**  
We need to solve  
$$
\sqrt{2(1-x)} = x-1.
$$  
**Step 1: Determine the domain.**  
- The square root requires  
  $$
  2(1-x) \ge 0 \quad \Longrightarrow \quad 1-x \ge 0 \quad \Longrightarrow \quad x \le 1.
  $$
- The square root gives a nonnegative number, so we must also have  
  $$
  x-1 \ge 0 \quad \Longrightarrow \quad x \ge 1.
  $$
Thus, $$x = 1$$ is the only candidate.

**Step 2: Verify $$x=1$$.**  
Substitute $$x=1$$ into the equation:  
$$
\sqrt{2(1-1)} = \sqrt{0} = 0 \quad \text{and} \quad 1-1 = 0.
$$  
The equation holds, so  
$$
x = 1.
$$

*Note:* Squaring both sides might give an extraneous solution (in this case $$x=-1$$), but it does not satisfy the original domain constraints.

---

**1.2**  
We need to solve for $$x$$ and $$y$$ simultaneously:  
$$
3^{x+y} = 27 \quad \text{and} \quad x^2 + y^2 = 17.
$$

**Step 1:**  
Express 27 as a power of 3:
$$
27 = 3^3.
$$
So,
$$
3^{x+y} = 3^3 \quad \Longrightarrow \quad x+y=3.
$$

**Step 2:** Substitute $$y = 3-x$$ into the second equation:
$$
x^2 + (3-x)^2 = 17.
$$
Expanding,
$$
x^2 + 9 - 6x + x^2 = 17 \quad \Longrightarrow \quad 2x^2 - 6x + 9 = 17.
$$
Subtract 17 from both sides:
$$
2x^2 - 6x - 8 = 0.
$$
Divide the equation by 2:
$$
x^2 - 3x - 4 = 0.
$$
Factor or use the quadratic formula. The discriminant is
$$
\Delta = (-3)^2 - 4\cdot1\cdot(-4) = 9+16=25.
$$
Thus,
$$
x = \frac{3\pm \sqrt{25}}{2} = \frac{3\pm5}{2}.
$$
This gives:
$$
x = \frac{3+5}{2} = 4 \quad \text{or} \quad x = \frac{3-5}{2} = -1.
$$
Now, find $$y$$ using $$x+y=3$$:  
- If $$x=4$$, then $$y = 3-4 = -1$$.  
- If $$x=-1$$, then $$y = 3-(-1) = 4$$.

Thus, the solutions are:
$$
(x,y) = (4,-1) \quad \text{or} \quad (x,y) = (-1,4).
$$
**1.1.1** $$ x = 3 $$ or $$ x = 4 $$ **1.1.2** $$ x \approx 0.18 $$ or $$ x \approx -1.85 $$ **1.1.3** $$ -5 < x < 3 $$ **1.1.4** $$ x = 1 $$ **1.2** $$ (x, y) = (4, -1) $$ or $$ (x, y) = (-1, 4) $$

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