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Update Unit 2: Quadratic Equations 1.md

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James Su 2020-01-13 13:48:31 +00:00
parent 700d29c843
commit 2ec67c3ea4

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@ -25,15 +25,19 @@ Let $`h`$ be the height and $`b`$ be the base. $`h = 2b + 4`$.
$`\therefore (2b+4)(b) = 168(2)`$
$`2b^2 + 8b = 168(2)`$
$`2b^2 + 4b = 168(2)`$
$`b^2 + 4b - 168 = 0`$
$`b^2 + 2b - 168 = 0`$
$`b = \dfrac{-4 \pm \sqrt{16+4(168)}}{2}`$
$`b = \dfrac{-2 \pm \sqrt{4+4(168)}}{2}`$
$`b = \dfrac{-4 \pm 4\sqrt{43}}{2}`$
$`b = \dfrac{-2 \pm 26}{2}`$
$`b = -2 + 2\sqrt{85}`$
$`b = -1 \pm 26 \implies b = 25, -27`$
$`\because b \gt 0 `$
$`\therefore b = 25cm`$
### Question 2 a)