TOPIC: Algebra Sub-topic: Solve Quadratic equations by factorisation Solve for \( x \) : One a) \( 3 x^{2}-6 x=0 \) b) \( x^{2}-81=0 \) c) \( x^{2}-x=12 \) d) \( x(x-9)=-14 \) e) \( 3 x^{\frac{2}{3}}-7 x^{\frac{1}{3}}+2=0 \) f) \( x-3 x^{\frac{1}{2}}-4=0 \)

Sure! Let's solve each of the quadratic equations by factorization step by step. --- ### **a) \( 3x^{2} - 6x = 0 \)** **Step 1: Factor out the greatest common factor (GCF).** The GCF of \( 3x^{2} \) and \( -6x \) is \( 3x \). \[ 3x^{2} - 6x = 3x(x - 2) = 0 \] **Step 2: Apply the Zero Product Property.** If \( AB = 0 \), then either \( A = 0 \) or \( B = 0 \). \[ 3x = 0 \quad \text{or} \quad x - 2 = 0 \] **Step 3: Solve for \( x \).** \[ x = 0 \quad \text{or} \quad x = 2 \] **Solution:** \( x = 0 \) and \( x = 2 \) --- ### **b) \( x^{2} - 81 = 0 \)** **Step 1: Recognize the difference of squares.** \[ x^{2} - 81 = x^{2} - 9^{2} = (x - 9)(x + 9) = 0 \] **Step 2: Apply the Zero Product Property.** \[ x - 9 = 0 \quad \text{or} \quad x + 9 = 0 \] **Step 3: Solve for \( x \).** \[ x = 9 \quad \text{or} \quad x = -9 \] **Solution:** \( x = 9 \) and \( x = -9 \) --- ### **c) \( x^{2} - x = 12 \)** **Step 1: Bring all terms to one side to set the equation to zero.** \[ x^{2} - x - 12 = 0 \] **Step 2: Factor the quadratic equation.** We need two numbers that multiply to \(-12\) and add to \(-1\). \[ (x - 4)(x + 3) = 0 \] **Step 3: Apply the Zero Product Property.** \[ x - 4 = 0 \quad \text{or} \quad x + 3 = 0 \] **Step 4: Solve for \( x \).** \[ x = 4 \quad \text{or} \quad x = -3 \] **Solution:** \( x = 4 \) and \( x = -3 \) --- ### **d) \( x(x - 9) = -14 \)** **Step 1: Expand and bring all terms to one side.** \[ x^{2} - 9x + 14 = 0 \] **Step 2: Factor the quadratic equation.** We need two numbers that multiply to \(+14\) and add to \(-9\). \[ (x - 7)(x - 2) = 0 \] **Step 3: Apply the Zero Product Property.** \[ x - 7 = 0 \quad \text{or} \quad x - 2 = 0 \] **Step 4: Solve for \( x \).** \[ x = 7 \quad \text{or} \quad x = 2 \] **Solution:** \( x = 7 \) and \( x = 2 \) --- ### **e) \( 3x^{\frac{2}{3}} - 7x^{\frac{1}{3}} + 2 = 0 \)** **Step 1: Make a substitution to simplify the equation.** Let \( y = x^{\frac{1}{3}} \). Then \( y^{2} = x^{\frac{2}{3}} \). Substitute into the equation: \[ 3y^{2} - 7y + 2 = 0 \] **Step 2: Factor the quadratic in \( y \).** We need two numbers that multiply to \(3 \times 2 = 6\) and add to \(-7\). \[ 3y^{2} - 6y - y + 2 = 0 \\ 3y(y - 2) -1(y - 2) = (3y - 1)(y - 2) = 0 \] **Step 3: Apply the Zero Product Property.** \[ 3y - 1 = 0 \quad \text{or} \quad y - 2 = 0 \] **Step 4: Solve for \( y \).** \[ y = \frac{1}{3} \quad \text{or} \quad y = 2 \] **Step 5: Substitute back \( y = x^{\frac{1}{3}} \) and solve for \( x \).** - For \( y = \frac{1}{3} \): \[ x^{\frac{1}{3}} = \frac{1}{3} \\ x = \left( \frac{1}{3} \right)^{3} = \frac{1}{27} \] - For \( y = 2 \): \[ x^{\frac{1}{3}} = 2 \\ x = 2^{3} = 8 \] **Solution:** \( x = \frac{1}{27} \) and \( x = 8 \) --- ### **f) \( x - 3x^{\frac{1}{2}} - 4 = 0 \)** **Step 1: Make a substitution to simplify the equation.** Let \( y = x^{\frac{1}{2}} \) (which means \( y = \sqrt{x} \)). Then \( y^{2} = x \). Substitute into the equation: \[ y^{2} - 3y - 4 = 0 \] **Step 2: Factor the quadratic in \( y \).** We need two numbers that multiply to \(-4\) and add to \(-3\). \[ (y - 4)(y + 1) = 0 \] **Step 3: Apply the Zero Product Property.** \[ y - 4 = 0 \quad \text{or} \quad y + 1 = 0 \] **Step 4: Solve for \( y \).** \[ y = 4 \quad \text{or} \quad y = -1 \] **Step 5: Substitute back \( y = \sqrt{x} \) and solve for \( x \).** - For \( y = 4 \): \[ \sqrt{x} = 4 \\ x = 16 \] - For \( y = -1 \): \[ \sqrt{x} = -1 \] Since \( \sqrt{x} \) represents the principal (non-negative) square root, \( \sqrt{x} = -1 \) has no real solution. **Solution:** \( x = 16 \) --- **Summary of Solutions:** - **a)** \( x = 0 \) and \( x = 2 \) - **b)** \( x = 9 \) and \( x = -9 \) - **c)** \( x = 4 \) and \( x = -3 \) - **d)** \( x = 7 \) and \( x = 2 \) - **e)** \( x = \frac{1}{27} \) and \( x = 8 \) - **f)** \( x = 16 \)