Exercise: 1. Solve for \( x \) : \( 1.12 x(x-3)=0 \) \( 1.23 x^{2}-5 x-4=0 \) \( 1.3 \sqrt{2-7 x}+2=x \) \( 1.4 \frac{1}{x+1}+\frac{2 x}{x-1}=1 \) \( 1.5 \frac{2}{3} x^{2}-2 x=6 \) \( 1.6(2 x-5)(x+3)=5 \) \( 1.7 x(2 x+5) \geq 3 \) \( 1.8(x-2)(3-x)>0 \) \( 1.9(5-x)^{2} \leq 0 \)

**1.12** We have \[ x(x-3)=0. \] So, either \[ x=0\quad \text{or}\quad x-3=0\implies x=3. \] Thus, the solutions are \(\;x=0\) and \(\;x=3\). --- **1.23** The equation is \[ 1.23\,x^{2}-5x-4=0. \] Using the quadratic formula with \(a=1.23\), \(b=-5\), and \(c=-4\), \[ x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}=\frac{5\pm\sqrt{(-5)^{2}-4\cdot1.23\cdot(-4)}}{2\cdot1.23}. \] Compute the discriminant: \[ (-5)^2-4\cdot1.23\cdot(-4)=25+19.68=44.68. \] Thus, \[ x=\frac{5\pm\sqrt{44.68}}{2.46}. \] Approximately, \[ \sqrt{44.68}\approx6.683, \] so the approximate solutions are \[ x\approx\frac{5+6.683}{2.46}\approx4.75,\quad x\approx\frac{5-6.683}{2.46}\approx-0.68. \] --- **1.3** The equation is \[ 1.3\sqrt{2-7x}+2=x. \] **Step 1.** Isolate the square root: \[ 1.3\sqrt{2-7x}=x-2. \] **Domain restriction:** The square root is defined if \[ 2-7x\ge0\quad\Longrightarrow\quad x\le\frac{2}{7}. \] Also, since the left side is nonnegative, we must have \[ x-2\ge0\quad\Longrightarrow\quad x\ge2. \] These two conditions (\(x\le\frac{2}{7}\) and \(x\ge2\)) cannot be satisfied simultaneously. Thus, there is no solution. --- **1.4** The equation is \[ \frac{1}{x+1}+\frac{2x}{x-1}=1. \] **Step 1.** Multiply both sides by \((x+1)(x-1)\) (noting that \(x\neq-1\) and \(x\neq1\)): \[ (x-1)+2x(x+1)=(x+1)(x-1). \] **Step 2.** Expand each term: \[ x-1+2x^2+2x=x^2-1. \] **Step 3.** Combine like terms: \[ 2x^2+3x-1=x^2-1. \] Subtract \(x^2-1\) from both sides: \[ 2x^2+3x-1-x^2+1=0\quad\Longrightarrow\quad x^2+3x=0. \] **Step 4.** Factor: \[ x(x+3)=0. \] So, either \(x=0\) or \(x=-3\). Checking the restrictions, for \(x=0\): \(0+1\neq0\) and \(0-1\neq0\). For \(x=-3\): \(-3+1=-2\neq0\) and \(-3-1=-4\neq0\). Thus, the solutions are \(\;x=0\) and \(\;x=-3\). --- **1.5** The equation is \[ \frac{2}{3}x^{2}-2x=6. \] **Step 1.** Multiply both sides by 3 to eliminate fractions: \[ 2x^2-6x=18. \] **Step 2.** Bring all terms to one side: \[ 2x^2-6x-18=0. \] Divide by 2: \[ x^2-3x-9=0. \] **Step 3.** Apply the quadratic formula with \(a=1\), \(b=-3\), \(c=-9\): \[ x=\frac{-(-3)\pm\sqrt{(-3)^2-4\cdot1\cdot(-9)}}{2\cdot1}=\frac{3\pm\sqrt{9+36}}{2}=\frac{3\pm\sqrt{45}}{2}. \] Since \(\sqrt{45}=3\sqrt{5}\), the solutions are \[ x=\frac{3\pm3\sqrt{5}}{2}. \] --- **1.6** The equation is \[ (2x-5)(x+3)=5. \] **Step 1.** Expand the left side: \[ 2x^2+6x-5x-15=2x^2+x-15. \] So the equation becomes: \[ 2x^2+x-15=5. \] **Step 2.** Bring all terms to one side: \[ 2x^2+x-20=0. \] **Step 3.** Use the quadratic formula with \(a=2\), \(b=1\), \(c=-20\): \[ x=\frac{-1\pm\sqrt{1^2-4\cdot2\cdot(-20)}}{2\cdot2}=\frac{-1\pm\sqrt{1+160}}{4}=\frac{-1\pm\sqrt{161}}{4}. \] --- **1.7** The inequality is \[ x(2x+5)\ge3. \] **Step 1.** Expand the left side: \[ 2x^2+5x\ge3. \] **Step 2.** Bring all terms to one side: \[ 2x^2+5x-3\ge0. \] **Step 3.** Find the roots of the quadratic equation \(2x^2+5x-3=0\). Use the quadratic formula with \(a=2\), \(b=5\), and \(c=-3\): \[ x=\frac{-5\pm\sqrt{5^2-4\cdot2\cdot(-3)}}{2\cdot2}=\frac{-5\pm\sqrt{25+24}}{4}=\frac{-5\pm\sqrt{49}}{4}. \] Thus, \[ x=\frac{-5+7}{4}=\frac{2}{4}=0.5,\quad x=\frac{-5-7}{4}=-3. \] **Step 4.** Since the quadratic opens upward (as \(a>0\)), the inequality holds when \[ x\le -3 \quad \text{or} \quad x\ge 0.5. \] --- **1.8** The inequality is \[ (x-2)(3-x)>0. \] **Step 1.** Note that \[ 3-x=-(x-3), \] so the inequality becomes \[ -(x-2)(x-3)>0 \quad\Longrightarrow\quad (x-2)(x-3)<0. \] **Step 2.** The product \((x-2)(x-3)\) is negative when \(x\) is strictly between the two roots. Thus, the inequality holds for \[ 2