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