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Update Quadratic Equations.md

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James Su 2019-12-30 02:35:44 +00:00
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Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Quadratic Equations.md
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@ -121,6 +121,7 @@ $`\therefore`$ The $`x`$-intercepts are at $`\dfrac{2}{3}, \dfrac{-1}{4}`$
### Question 3 c) ### Question 3 c)
When $`P(x) = 0`$, that means it is the break-even point for a value of $`x`$ (no profit, no loss). When $`P(x) = 0`$, that means it is the break-even point for a value of $`x`$ (no profit, no loss).
$`2k^2 + 12k - 10 = 0 \implies k^2 -6k + 5 = 0`$ $`2k^2 + 12k - 10 = 0 \implies k^2 -6k + 5 = 0`$
$`(k-5)(k-1) = 0`$ $`(k-5)(k-1) = 0`$
@ -158,7 +159,7 @@ $`x = \dfrac{3 \pm \sqrt{23}}{7}`$
$`\because \dfrac{1}{3}`$ 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 $`\because \dfrac{1}{3}`$ 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. gives those roots.
After expanind we get: After expanding we get:
$`y = x^2 - \dfrac{2}{3}x - \dfrac{1}{3}x + \dfrac{2}{3}`$ $`y = x^2 - \dfrac{2}{3}x - \dfrac{1}{3}x + \dfrac{2}{3}`$
@ -238,15 +239,10 @@ Since we know that $`(4, 5)`$ is also a point on the parabola, we can susbsitute
$`5 = a + c \quad (2)`$ $`5 = a + c \quad (2)`$
```math ```math
\begin{cases} \begin{cases}
9a + c = 0 & \text{(1)} \\ 9a + c = 0 & \text{(1)} \\
a + c = 5 & \text{(2)} \\ a + c = 5 & \text{(2)} \\
\end{cases} \end{cases}
``` ```
$`(2) - (1)`$ $`(2) - (1)`$