1. **Problem 1.1: Solve the quadratic equation** $$x^2 + 14x + 45 = 0$$ Step 1: Identify coefficients: $$a=1$$, $$b=14$$, and $$c=45$$. Step 2: Calculate the discriminant $$\Delta$$ using $$\Delta = b^2 - 4ac$$: $$\Delta = 14^2 - 4\times1\times45 = 196 - 180 = 16$$. Step 3: Since $$\Delta > 0$$, the equation has two real roots. Step 4: Use quadratic formula $$x = \frac{-b \pm \sqrt{\Delta}}{2a}$$: $$x = \frac{-14 \pm \sqrt{16}}{2} = \frac{-14 \pm 4}{2}$$. Step 5: Calculate roots: - $$x_1 = \frac{-14 + 4}{2} = \frac{-10}{2} = -5$$ - $$x_2 = \frac{-14 - 4}{2} = \frac{-18}{2} = -9$$ **Answer:** $$x = -5$$ or $$x = -9$$. 2. **Problem 1.2: Factorize the expressions** a) $$x^2 + 12x + 36$$ Step 1: Recognize this as a perfect square trinomial: $$(x + 6)^2$$ because $$6 \times 6 = 36$$ and $$6 + 6 = 12$$. **Answer:** $$(x + 6)^2$$ b) $$x^2 - 25$$ Step 1: Recognize it as a difference of squares: $$x^2 - 25 = (x)^2 - (5)^2$$ Step 2: Apply difference of squares factorization: $$(x - 5)(x + 5)$$ **Answer:** $$(x - 5)(x + 5)$$ 3. **Problem 1.3: Rectangle dimensions given area and length-width relation** Step 1: Let width be $$w$$ cm, then length is $$w + 2$$ cm. Step 2: Area $$A = \text{length} \times \text{width} = (w + 2)w = 8$$ Step 3: Form quadratic equation: $$w^2 + 2w = 8$$ Rewrite as: $$w^2 + 2w - 8 = 0$$ Step 4: Factorize or use quadratic formula: Calculate discriminant: $$\Delta = 2^2 - 4\times1\times(-8) = 4 + 32 = 36$$ Step 5: Use quadratic formula: $$w = \frac{-2 \pm \sqrt{36}}{2} = \frac{-2 \pm 6}{2}$$ Step 6: Calculate roots: - $$w_1 = \frac{-2 + 6}{2} = \frac{4}{2} = 2$$ - $$w_2 = \frac{-2 - 6}{2} = \frac{-8}{2} = -4$$ (discard negative width) Step 7: Length is $$w + 2 = 2 + 2 = 4$$ cm. **Answer:** Width = 2 cm, Length = 4 cm.