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