2.5 KiB
Quadratic Equations 2
Question 4 a)
\(`5x(x-1) + 5 = 7 + x(1-2x)`\)
\(`5x^2 - 5x = 2 + x - 2x^2`\)
\(`7x^2 - 6x - 2 = 0`\)
\(`x = \dfrac{6 \pm \sqrt{92}}{14} = \dfrac{3 \pm \sqrt{23}}{7}`\)
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. Here we make \(`a=1`\), so its easy to find a quadratic in vertex from that gives these roots.
Let the vertex form then be \(`y=(x-d)^2+c`\), since \(`a=1`\).
We know \(`d=\dfrac{r_1+r_2}{2}`\) because it is the x-coordinate of the vertex which is also the AOS. Therefore it is equal to \(`\dfrac{\dfrac{1}{3} + \dfrac{-2}{3}}{2} = \dfrac{-1}{6}`\)
Then, we know \(`c=(d-\dfrac{1}{3})(d+\dfrac{2}{3})`\), since by plugging in the x-coordinate of the vertex, we get the y-coordinate of the vertex which is also the \(`c`\) value.
Therefore \(`c=(\dfrac{-1}{6}-\dfrac{1}{3})(\dfrac{-1}{6} + \dfrac{2}{3}) = \dfrac{-1}{4}`\).
Therefore our equation is simply \(`y = (x+\dfrac{1}{6})^2 - \dfrac{1}{4}`\)
Qustion 4 c)
When \(`h = 0`\), the ball hits ground, so:
\(`-3.2t^2 + 12.8 + 1 = 0`\)
\(`t = \dfrac{12.8 \pm \sqrt{176.64}}{6.4}`\)
\(`\because t \ge 0`\)
\(`\therefore t \approx 4.1`\)
The ball will strike the ground at approximately \(`4.1`\) seconds.
Question 5 a)
\(`128 = 96t - 16t^2`\)
\(`16t^2 - 96t + 128 = 0`\)
\(`t^2 - 6t + 8 = 0`\)
\(`(t-2)(t-4)`\), at seconds \(`2`\) and \(`4`\), the rocket reaches \(`128m`\).
Question 5 b)
Break even is when revenue = cost.
\(`\therefore R(d) = C(d)`\)
\(`-40d^2 + 200d = 300 - 40d`\)
\(`40d^2 - 240d + 300 = 0`\)
\(`2d^2 - 12d + 15 = 0`\)
\(`d = \dfrac{12 \pm \sqrt{24}}{4}`\)
\(`d = \dfrac{6 \pm \sqrt{6}}{2}`\)
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 \(`y = \dfrac{-5}{8}(x-3)^2 + \dfrac{45}{8}`\)