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Update Quadratic Equations.md

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James Su 2019-12-30 02:34:07 +00:00
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Grade 10/Math/MPM2DZ/Math Oral Presentation Questions/Quadratic Equations.md
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@ -78,10 +78,6 @@ $`4w^2 + 30w = 54`$
$`2w^2 + 15w - 27 = 0`$ $`2w^2 + 15w - 27 = 0`$
$`2 \quad \quad -3`$
$`1 \quad \quad 9`$
$`(2w-3)(w+9) = 0`$ $`(2w-3)(w+9) = 0`$
$`w = \dfrac{3}{2}, -9`$ $`w = \dfrac{3}{2}, -9`$
@ -118,10 +114,6 @@ Use discriminant, where $`D = b^2 - 4ac`$.
$`y = 12x^2 - 5x - 2`$ $`y = 12x^2 - 5x - 2`$
$`3 \quad \quad -2`$
$`4 \quad \quad 1`$
$`y = (3x-2)(4x+1)`$ $`y = (3x-2)(4x+1)`$
$`\therefore`$ The $`x`$-intercepts are at $`\dfrac{2}{3}, \dfrac{-1}{4}`$ $`\therefore`$ The $`x`$-intercepts are at $`\dfrac{2}{3}, \dfrac{-1}{4}`$
@ -192,7 +184,7 @@ $`t = \dfrac{12.8 \pm \sqrt{151.04}}{6.4}`$
$`\because t \ge 0`$ $`\because t \ge 0`$
$`\therefore t = \dfrac{12.8 \pm sqrt{151.04}}{6.4}`$ $`\therefore t = \dfrac{12.8 \pm \sqrt{151.04}}{6.4}`$
$`\therefore t \approx 3.9`$ $`\therefore t \approx 3.9`$
@ -206,10 +198,6 @@ $`16t^2 - 96t + 128 = 0`$
$`t^2 - 6t + 8 = 0`$ $`t^2 - 6t + 8 = 0`$
$`1 \quad \quad -2`$
$`1 \quad \quad -4`$
$`(t-2)(t-4)`$, at seconds $`2`$ and $`4`$, the rocket reaches $`128m`$. $`(t-2)(t-4)`$, at seconds $`2`$ and $`4`$, the rocket reaches $`128m`$.