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Update Quadratic Equations.md
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### Question 1 a)
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The zeroes of a quadratic equation are the solutiosn of $`x`$ when $`ax^2+bx+c = 0`$. The roots of the quadratic equation is when $`(x + r_1)(x + r_2) = 0`$, more
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commonly described by the formula $`\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}`$. Therefore the roots of a quadratic equation are also the zeroes of the quadratic equation.
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The zeroes of a quadratic equation are the solutions of $`x`$ when $`ax^2+bx+c = 0`$. The roots of the quadratic equation is when $`(x + r_1)(x + r_2) = 0`$, more
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commonly described by the formula $`\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}`$. Therefore the roots of a quadratic equation are also the zeroes of the quadratic equation.
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### Question 1 b)
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$`D = b^2 - 4ac`$
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If $`D=0`$, there is one zero.
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$`\therefore n^2 - 4(1)(4) = 0`$
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$`n^2 - 16 = 0`$
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$`(n+4)(n-4) = 0`$
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$`n = \pm 4`$
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### Question 1 c)
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Let $`h`$ be the height and $`b`$ be the base. $`h = 2b + 4`$.
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$`\therefore (2b+4)(b) = 168(2)`$
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$`4b^2 + 8b = 168(2)`$
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$`b^2 + 2b = 84 = 0`$
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$`b = \dfrac{-2 \pm \sqrt{4+4(84)}`$
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$`b = \dfrac{-2 \pmm \sqrt{340}}{2}`$
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$`b = -1 + \sqrt{85}`$
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### Question 2 a)
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If the quadratic is in $`ax^2 + bx + c`$, the AOS (axis of symmetry) is at $`\dfrac{-b}{2a}`$. And you can plug that value into the quadratic equation to get your optimal value,
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which is:
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$`= a(\dfrac{-b}{2a})^2 + b(\dfrac{-b}{2a}) + c`$
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$`= \dfrac{b^2}{4a} + \dfrac{-b^2}{2a} + c`$
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$`= \dfrac{-b^2}{4a} + c`$
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### Question 2 b)
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$`2x^2 + 5x - 1 = 0`$
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$`2(x^2 + \dfrac{5}{2}x + (\dfrac{5}{4})^2 - (\dfrac{5}{4})^2) - 1 = 0`$
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$`2(x+\dfrac{5}{4})^2 - \dfrac{25}{8} - 1 = 0`$
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$`2(x+ \dfrac{5}{4})^2 - \dfrac{33}{8} = 0`$
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$`(x+ \dfrac{5}{4})^2 = \dfrac{33}{16}`$
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$`x = \pm \sqrt{\dfrac{33}{16}} - \dfrac{5}{4}`$
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$`x = \dfrac{\pm \sqrt{33}}{4} - \dfrac{5}{4}`$
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$`x = \dfrac{\sqrt{33}-5}{4}`$ or $`\dfrac{-\sqrt{33} - 5}{4}`$
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### Question 2 c)
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Let $`w`$ be the width between the path and flowerbed, $`x`$ be the length of the whole rectangle and $`y`$ be the whole rectangle (flowerbed + path).
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$`x = 9+2w`$
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$`y = 6+2w`$
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$`(6+2w)(9+2w) - (6)(9 = (6)(9)`$
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$`54 + 12w + 18w + 4w^2 = 2(54)`$
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$`4w^2 + 30w = 54`$
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$`2w^2 + 15w - 27 = 0`$
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$`2 \quad \quad -3`$
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$`1 \quad \quad 9`$
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$`(2w-3)(w+9) = 0`$
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$`w = \dfrac{3}{2}, -9`$
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$`\because w \gt 0`$
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$`\therefore w = \dfrac{3}{2}`$
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$`\therefore x = 9+3 = 12`$
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$`\therefore y = 6+3 = 9`$
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$`P = 2(x + y) \implies P = 2(12+9) \implies P = 42`$
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$`\therefore`$ The perimeter is $`42m`$
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### Question 3 a)
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Use discriminant, where $`D = b^2 - 4ac`$.
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```math
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\begin{cases}
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\text{If } D \gt 0 & \text{Then there are 2 real distinct solutions} \\
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\text{If } D = 0 & \text{Then there is 1 real solution} \\
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\text{If } D \lt 0 &\text{Then there are no real solutions} \\
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\end{cases}
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```
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### Question 3 b)
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$`y = 12x^2 - 5x - 2`$
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$`3 \quad \quad -2`$
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$`4 \quad \quad 1`$
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$`y = (3x-2)(4x+1)`$
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$`\therefore`$ The $`x`$-intercepts are at $`\dfrac{2}{3}, \dfrac{-1}{4}`$
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### Question 3 c)
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When $`P(x) = 0`$, that means it is the break-even point for a value of $`x`$ (no profit, no loss).
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$`2k^2 + 12k - 10 = 0 \implies k^2 -6k + 5 = 0`$
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$`(k-5)(k-1) = 0`$
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$`k = 5, 1`$
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Either $`5000`$ or $`1000`$ rings must be produced so that there is no prodift and no less.
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AOS (axis of symmetry) = $`\dfrac{-b}{2a} = \dfrac{6}{2} = 3`$
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$`\therefore 3000`$ rings should be made to achieve the optimal value.
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Maximum profit $`= -2(3)^2 + 12(3) - 10`$
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$`= -18 + 30 - 10`$
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$` = 8`$
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$`\therefore 8000`$ dollars is the maximum profit.
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### Question 4 a)
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$`5x(x-1) + 5 = 7 + x(1-2x)`$
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$`5x^2 - 5x = 2 + x - 2x^2`$
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$`7x^2 - 6x - 2 = 0`$
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$`x = \dfrac{6 \pm \sqrt{92}}{14}`$
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$`x = \dfrac{3 \pm \sqrt{23}}{7}`$
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### Question 4 b).
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$`\because \dfrac{1}{3}`$ and $`\dfrac{-2}{3}`$ are the roots of a quadratic equation, that must mean that $`(x-\dfrac{1}{3})(x-\dfrac{2}{3})`$ is a quadratic equation that
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gives those roots.
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After expanind we get:
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$`y = x^2 - \dfrac{2}{3}x - \dfrac{1}{3}x + \dfrac{2}{3}`$
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$`y = x^2 - x + \dfrac{2}{3}`$
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Now we complete the square.
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$`y = x^2 - x + (\dfrac{1}{2})^2 - (\dfrac{1}{2})^2 + \dfrac{2}{3}`$
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$`y = (x-\dfrac{1}{2})^2 - \dfrac{1}{4} + \dfrac{2}{3}`$
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$`y = (x-\dfrac{1}{2})^2 + \dfrac{5}{12}`$
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### Qustion 4 c)
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When $`h = 0`$, the ball hits ground, so:
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$`-3.2t^2 + 12.8 + 1 = 0`$
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$`t = \dfrac{-12.8 \pm \sqrt{12.8^2 - 4(3.2)}}{-6.4}`$
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$`t = \dfrac{12.8 \pm \sqrt{151.04}}{6.4}`$
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$`\because t \ge 0`$
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$`\therefore t = \dfrac{12.8 \pm sqrt{151.04}}{6.4}`$
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$`\therefore t \approx 3.9`$
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The ball will strike the ground at approximately $`3.9`$ seconds.
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### Question 5 a)
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$`128 = 96t - 16t^2`$
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$`16t^2 - 96t + 128 = 0`$
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$`t^2 - 6t + 8 = 0`$
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$`1 \quad \quad -2`$
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$`1 \quad \quad -4`$
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$`(t-2)(t-4)`$, at seconds $`2`$ and $`4`$, the rocket reaches $`128m`$.
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### Question 5 b)
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Break even is when revenue = cost.
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$`\therefore R(d) = C(d)`$
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$`-40d^2 + 200d = 300 - 40d`$
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$`40d^2 - 240d + 300 = 0`$
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$`4d^2 - 24d + 30 = 0`$
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$`2d^2 - 12d + 15 = 0`$
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$`d = \dfrac{12 \pm \sqrt{24}}{4}`$
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$`d = \dfrac{6 \pm \sqrt{6}}{2}`$
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At $`d = 4.22474407$ or $`1.775255135`$ is when you break even.
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### Question 5 c)
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Let the quation be $`y = a(x-d)^2 + c`$
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Since we know that that $`(0, 0)`$ and $`(6, 0)`$ are the roots of this equation, the AOS is when $`x = 3`$
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$`\therefore y = a(x-3)^2 + c`$.
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Since we know that $`(0, 0)`$ is a point on the parabola, we can susbsitute it into our equation.
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$`0 = 9a + c \quad (1)`$
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Since we know that $`(4, 5)`$ is also a point on the parabola, we can susbsitute it int our equation as well.
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$`5 = a + c \quad (2)`$
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```math
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\begin{cases}
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9a + c = 0 & \text{(1)} \\
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a + c = 5 & \text{(2)} \\
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\end{cases}
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```
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$`(2) - (1)`$
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$`-8a = 5 \implies a = \dfrac{-5}{8} \quad (3)`$
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Sub $`3`$ into $`(2)`$:
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$`5 = \dfrac{-5}{8} + c \implies c = \dfrac{45}{8}`$
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$`\therefore`$ Our equation is $`y = \dfrac{-5}{8}(x-3)^2 + \dfrac{45}{8}`$
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