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Update Unit 2: Quadratic Equations.md

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James Su 2020-01-02 19:01:07 +00:00
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Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Unit 2: Quadratic Equations.md
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@ -9,13 +9,13 @@ $`D = b^2 - 4ac`$
If $`D=0`$, there is one zero.
$`\therefore n^2 - 4(1)(4) = 0`$
$`\therefore n^2 - 4(1)(9) = 0`$
$`n^2 - 16 = 0`$
$`n^2 - 36 = 0`$
$`(n+4)(n-4) = 0`$
$`(n+6)(n-6) = 0`$
$`n = \pm 4`$
$`n = \pm 6`$
### Question 1 c)
@ -23,15 +23,15 @@ Let $`h`$ be the height and $`b`$ be the base. $`h = 2b + 4`$.
$`\therefore (2b+4)(b) = 168(2)`$
$`4b^2 + 8b = 168(2)`$
$`2b^2 + 8b = 168(2)`$
$`b^2 + 2b = 84 = 0`$
$`b^2 + 4b - 168 = 0`$
$`b = \dfrac{-2 \pm \sqrt{4+4(84)}}{4}`$
$`b = \dfrac{-4 \pm \sqrt{16+4(168)}}{2}`$
$`b = \dfrac{-2 \pm \sqrt{340}}{2}`$
$`b = \dfrac{-4 \pm 4\sqrt{43}}{2}`$
$`b = -1 + \sqrt{85}`$
$`b = -2 + 2\sqrt{85}`$
### Question 2 a)
@ -154,22 +154,19 @@ $`x = \dfrac{6 \pm \sqrt{92}}{14} = \dfrac{3 \pm \sqrt{23}}{7}`$
### 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.
$`\because \dfrac{1}{3}`$ 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.
After expanding we get:
Let the vertex form then be $`y=(x-d)^2+c`$, since $`a=1`$.
$`y = x^2 - \dfrac{2}{3}x - \dfrac{1}{3}x + \dfrac{2}{3}`$
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 $`\dfrac{\dfrac{1}{3} + \dfrac{-2}{3}}{2} = \dfrac{-1}{6}`$
$`y = x^2 - x + \dfrac{2}{3}`$
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.
Now we complete the square.
Therefore $`c=(\dfrac{-1}{6}-\dfrac{1}{3})(\dfrac{-1}{6} + \dfrac{2}{3}) = \dfrac{-1}{4}`$.
$`y = x^2 - x + (\dfrac{1}{2})^2 - (\dfrac{1}{2})^2 + \dfrac{2}{3}`$
Therefore our equation is simply $`y = (x+\dfrac{1}{6})^2 - \dfrac{1}{4}`$
$`y = (x-\dfrac{1}{2})^2 - \dfrac{1}{4} + \dfrac{2}{3}`$
$`y = (x-\dfrac{1}{2})^2 + \dfrac{5}{12}`$
### Qustion 4 c)