From 29554e6656412f9d1f9e1c0911250d5ab6747d14 Mon Sep 17 00:00:00 2001
From: James Su <amagicalsoup@gmail.com>
Date: Thu, 2 Jan 2020 19:01:07 +0000
Subject: [PATCH] Update Unit 2: Quadratic Equations.md

---
 .../Unit 2: Quadratic Equations.md            | 35 +++++++++----------
 1 file changed, 16 insertions(+), 19 deletions(-)

diff --git a/Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Unit 2: Quadratic Equations.md b/Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Unit 2: Quadratic Equations.md
index 42cc9f9..e25b6c6 100644
--- a/Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Unit 2: Quadratic Equations.md	
+++ b/Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Unit 2: Quadratic Equations.md	
@@ -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)