$$ \left. \begin{array} { l l } { 1.1 .1 } & { x ^ { 2 } - 6 x = 0 } \ { 1.1 .2 } & { x ( x + 3 ) - 1 = 0 } \ { 1.1 .3 } & { \sqrt { x ^ { 3 } } = 512 } \ { 1.1 .4 } & { ( x - 1 ) ( x - 2 ) \leq 6 } \end{array} ight. $$

Question

$$ \left. \begin{array} { l l } { 1.1 .1 } & { x ^ { 2 } - 6 x = 0 } \ { 1.1 .2 } & { x ( x + 3 ) - 1 = 0 } \ { 1.1 .3 } & { \sqrt { x ^ { 3 } } = 512 } \ { 1.1 .4 } & { ( x - 1 ) ( x - 2 ) \leq 6 } \end{array} ight. $$

UpStudy AI · Accepted Answer

### 1.1.1 Solve $$ x^2 - 6x = 0 $$

1. Factor the left-hand side:
   $$
   x^2 - 6x = x(x - 6)
   $$
2. Set each factor equal to zero:
   $$
   x = 0 \quad \text{or} \quad x - 6 = 0
   $$
3. Solve for $$ x $$:
   $$
   x = 0 \quad \text{or} \quad x = 6
   $$

---

### 1.1.2 Solve $$ x(x+3) - 1 = 0 $$

1. Expand the expression:
   $$
   x^2 + 3x - 1 = 0
   $$
2. Use the quadratic formula:
   $$
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   $$
   Here, $$ a = 1 $$, $$ b = 3 $$, and $$ c = -1 $$.

3. Calculate the discriminant:
   $$
   \Delta = 3^2 - 4(1)(-1) = 9 + 4 = 13
   $$
4. Determine the solutions:
   $$
   x = \frac{-3 \pm \sqrt{13}}{2}
   $$

---

### 1.1.3 Solve $$\sqrt{x^3} = 512$$

1. Write the equation in exponent form. Since $$\sqrt{x^3} = (x^3)^{\frac{1}{2}} = x^{\frac{3}{2}}$$:
   $$
   x^{\frac{3}{2}} = 512
   $$
2. Solve for $$x$$ by raising both sides to the power of $$\frac{2}{3}$$:
   $$
   x = 512^{\frac{2}{3}}
   $$
3. Recognize that $$512 = 8^3$$ and compute:
   $$
   512^{\frac{2}{3}} = \left(8^3\right)^{\frac{2}{3}} = 8^2 = 64
   $$

---

### 1.1.4 Solve $$(x-1)(x-2) \leq 6$$

1. Expand the left-hand side:
   $$
   (x-1)(x-2) = x^2 - 3x + 2
   $$
2. Write the inequality:
   $$
   x^2 - 3x + 2 \leq 6
   $$
3. Subtract $$6$$ from both sides:
   $$
   x^2 - 3x + 2 - 6 \leq 0 \quad \Longrightarrow \quad x^2 - 3x - 4 \leq 0
   $$
4. Factor the quadratic or use the quadratic formula to find its roots. The quadratic equation is:
   $$
   x^2 - 3x - 4 = 0
   $$
5. Compute the discriminant:
   $$
   \Delta = (-3)^2 - 4(1)(-4) = 9 + 16 = 25
   $$
6. Find the roots:
   $$
   x = \frac{3 \pm \sqrt{25}}{2} = \frac{3 \pm 5}{2}
   $$
   Thus,
   $$
   x = \frac{8}{2} = 4 \quad \text{or} \quad x = \frac{-2}{2} = -1
   $$
7. Since the quadratic opens upward ($$a > 0$$), the inequality $$x^2 - 3x - 4 \leq 0$$ holds between the roots:
   $$
   -1 \leq x \leq 4
   $$
1.1.1: $$ x = 0 $$ or $$ x = 6 $$ 1.1.2: $$ x = \frac{-3 + \sqrt{13}}{2} $$ or $$ x = \frac{-3 - \sqrt{13}}{2} $$ 1.1.3: $$ x = 64 $$ 1.1.4: $$ -1 \leq x \leq 4 $$

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