1. Use the discriminant to determine the number of solutions for each quadratic. The discriminant formula is $$\Delta = b^2 - 4ac$$ where the quadratic is $$ax^2 + bx + c = 0$$. (a) For $$x^2 + 5x - 3 = 0$$, $$a=1, b=5, c=-3$$. Calculate $$\Delta = 5^2 - 4(1)(-3) = 25 + 12 = 37$$. Since $$\Delta > 0$$, there are 2 distinct real solutions. (b) For $$4x^2 + 4x + 1 = 0$$, $$a=4, b=4, c=1$$. Calculate $$\Delta = 4^2 - 4(4)(1) = 16 - 16 = 0$$. Since $$\Delta = 0$$, there is exactly 1 real solution (a repeated root). (c) For $$2x^2 + 5x + 4 = 0$$, $$a=2, b=5, c=4$$. Calculate $$\Delta = 5^2 - 4(2)(4) = 25 - 32 = -7$$. Since $$\Delta < 0$$, there are no real solutions (2 complex solutions). 2. Express $$x^2 + 2x + 3$$ in the form $$(x + a)^2 + b$$. (a) Complete the square: $$x^2 + 2x + 3 = (x^2 + 2x + 1) + 3 - 1 = (x + 1)^2 + 2$$. So, $$a = 1$$ and $$b = 2$$. (b) Sketch the graph of $$y = x^2 + 2x + 3$$. - Vertex is at $$(-a, b) = (-1, 2)$$. - Find y-intercept by setting $$x=0$$: $$y = 0 + 0 + 3 = 3$$. - Find x-intercepts by solving $$x^2 + 2x + 3 = 0$$, which has no real roots (from discriminant). So, the graph is a parabola opening upwards with vertex at $$(-1, 2)$$, crossing y-axis at $$(0,3)$$, and no x-intercepts. (c) Discriminant of $$x^2 + 2x + 3$$ is $$\Delta = 2^2 - 4(1)(3) = 4 - 12 = -8$$. Since $$\Delta < 0$$, no real roots exist, consistent with the graph having no x-intercepts. 3. For $$4x^2 + 12x = k$$ to have equal roots, discriminant must be zero. Rewrite as $$4x^2 + 12x - k = 0$$. Here, $$a=4, b=12, c=-k$$. Set discriminant to zero: $$12^2 - 4(4)(-k) = 0 \Rightarrow 144 + 16k = 0 \Rightarrow 16k = -144 \Rightarrow k = -9$$. 4. Given $$y = 2^x$$: (a) Express $$4^x$$ in terms of $$y$$. Note $$4^x = (2^2)^x = 2^{2x} = (2^x)^2 = y^2$$. (b) Solve $$8(4^x) - 9(2^x) + 1 = 0$$. Substitute $$4^x = y^2$$ and $$2^x = y$$: $$8y^2 - 9y + 1 = 0$$. Solve quadratic in $$y$$: Discriminant $$= (-9)^2 - 4(8)(1) = 81 - 32 = 49$$. Roots: $$y = \frac{9 \pm 7}{16}$$. So, $$y_1 = \frac{16}{16} = 1$$, $$y_2 = \frac{2}{16} = \frac{1}{8}$$. Recall $$y = 2^x$$. For $$y=1$$, $$2^x = 1 \Rightarrow x=0$$. For $$y=\frac{1}{8}$$, $$2^x = 2^{-3} \Rightarrow x = -3$$. 5. For $$x^2 + (3k + 1)x - k = 0$$ to have equal roots, discriminant zero. Here, $$a=1, b=3k+1, c=-k$$. Set discriminant zero: $$(3k+1)^2 - 4(1)(-k) = 0 \Rightarrow (3k+1)^2 + 4k = 0$$. Expand: $$9k^2 + 6k + 1 + 4k = 0 \Rightarrow 9k^2 + 10k + 1 = 0$$. Solve quadratic for $$k$$: Discriminant: $$10^2 - 4(9)(1) = 100 - 36 = 64$$. Roots: $$k = \frac{-10 \pm 8}{18}$$. So, $$k_1 = \frac{-10 + 8}{18} = \frac{-2}{18} = -\frac{1}{9}$$, $$k_2 = \frac{-10 - 8}{18} = \frac{-18}{18} = -1$$. 6. Solve $$x - \frac{5}{x} = 4$$. Multiply both sides by $$x$$ (assuming $$x \neq 0$$): $$x^2 - 5 = 4x$$. Rewrite: $$x^2 - 4x - 5 = 0$$. Solve quadratic: Discriminant: $$(-4)^2 - 4(1)(-5) = 16 + 20 = 36$$. Roots: $$x = \frac{4 \pm 6}{2}$$. So, $$x_1 = \frac{10}{2} = 5$$, $$x_2 = \frac{-2}{2} = -1$$. 7. The line $$y = 3x - 7$$ is tangent to $$y = kx^2 + 2x - 1$$. Set equal: $$kx^2 + 2x - 1 = 3x - 7$$. Rewrite: $$kx^2 + 2x - 1 - 3x + 7 = 0 \Rightarrow kx^2 - x + 6 = 0$$. For tangency, discriminant zero: $$(-1)^2 - 4(k)(6) = 0 \Rightarrow 1 - 24k = 0 \Rightarrow 24k = 1 \Rightarrow k = \frac{1}{24}$$. 8. Solve $$2x^{2/3} + x^{1/3} - 6 = 0$$. Let $$u = x^{1/3}$$, then $$u^2 = x^{2/3}$$. Rewrite: $$2u^2 + u - 6 = 0$$. Solve quadratic: Discriminant: $$1^2 - 4(2)(-6) = 1 + 48 = 49$$. Roots: $$u = \frac{-1 \pm 7}{4}$$. So, $$u_1 = \frac{6}{4} = 1.5$$, $$u_2 = \frac{-8}{4} = -2$$. Recall $$u = x^{1/3}$$. Cube both sides: $$x_1 = (1.5)^3 = 3.375$$, $$x_2 = (-2)^3 = -8$$. 9. Ball height model: $$h = 3 + 14t - 5t^2$$. (a) Height at $$t=0$$: $$h = 3 + 14(0) - 5(0)^2 = 3$$. (b) Time when ball hits floor: set $$h=0$$. $$3 + 14t - 5t^2 = 0$$. Rewrite: $$-5t^2 + 14t + 3 = 0$$ or $$5t^2 - 14t - 3 = 0$$. Solve quadratic: Discriminant: $$(-14)^2 - 4(5)(-3) = 196 + 60 = 256$$. Roots: $$t = \frac{14 \pm 16}{10}$$. So, $$t_1 = \frac{30}{10} = 3$$, $$t_2 = \frac{-2}{10} = -0.2$$ (discard negative time). Time ball hits floor is $$t=3$$ seconds. (c) Maximum height at vertex: Vertex time: $$t = -\frac{b}{2a} = -\frac{14}{2(-5)} = \frac{14}{10} = 1.4$$. Height at $$t=1.4$$: $$h = 3 + 14(1.4) - 5(1.4)^2 = 3 + 19.6 - 5(1.96) = 3 + 19.6 - 9.8 = 12.8$$. Maximum height is 12.8 meters. 10. Garden area and fence length. (i) Area $$A$$ is sum of rectangle and triangle areas. Rectangle area: $$x(x+4) = x^2 + 4x$$. Triangle area: $$\frac{1}{2} \times x \times (x+2) = \frac{x(x+2)}{2} = \frac{x^2 + 2x}{2}$$. Total area: $$A = x^2 + 4x + \frac{x^2 + 2x}{2} = x^2 + 4x + \frac{x^2}{2} + x = \frac{3x^2}{2} + 5x$$. (ii) Fence length $$P$$ is perimeter of combined shape. Fence includes: - Rectangle bottom: $$x + 4$$ - Rectangle left side: $$x$$ - Triangle right vertical side: $$x$$ - Triangle base: $$x + 2$$ - Hypotenuse of triangle: $$\sqrt{x^2 + (x+2)^2} = \sqrt{x^2 + x^2 + 4x + 4} = \sqrt{2x^2 + 4x + 4}$$. Total fence length: $$P = (x + 4) + x + x + (x + 2) + \sqrt{2x^2 + 4x + 4} = 4x + 6 + \sqrt{2x^2 + 4x + 4}$$. (iii) Given $$A = 200$$: $$\frac{3x^2}{2} + 5x = 200$$. Multiply both sides by 2: $$3x^2 + 10x = 400$$. Rewrite: $$3x^2 + 10x - 400 = 0$$. Solve quadratic: Discriminant: $$10^2 - 4(3)(-400) = 100 + 4800 = 4900$$. Roots: $$x = \frac{-10 \pm 70}{6}$$. Positive root: $$x = \frac{60}{6} = 10$$. Calculate fence length: $$P = 4(10) + 6 + \sqrt{2(10)^2 + 4(10) + 4} = 40 + 6 + \sqrt{200 + 40 + 4} = 46 + \sqrt{244}$$. Approximate: $$\sqrt{244} \approx 15.62$$. So, $$P \approx 46 + 15.62 = 61.62$$ meters. Final answer: Fence length is approximately 61.6 meters to 3 significant figures.