vault backup: 2024-01-03 10:02:43

education/math/Quadratics.md

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 # Notes
 
-To convert an equation in the form of $(x^2-3x)$ into a square equivalent, you half the second value, then square that value (in this case 3) $(x-\frac{3}{2})^2$, resulting in an equation like this. 
+To convert an equation in the form of $y = (x^2-3x)$ into a square equivalent, you:
+- Take the second value, $3$, then half it, giving you ($\frac{3}{2}$). 
+- Then to rebalance the equation, you're going to square that value (giving you $\frac{9}{4}$), then add it to the other side, multiplying it by $a$ if necessary, where $a$ is what the parentheses are multiplied by in the form $a(x - h)^2$. This will give you an equation that looks something like: $\frac{9}{4} + y = (x^2-3x)$.
+- Finally, you can rebalance the equation by subtracting $\fr
+
 $$ y = -5x^2 -20x + 13 $$
 Given the above equation, you can factor out a -5, resulting in the equation $-5(x^2+4x) + 13)$. Half of 4 is 2, and because the inside is multiplied by -5, $-5 *4 = -20$, so you add -20 to the other side to equalize the equation, resulting in an equation in the form of $-20 + y = -5(x+2)^2+ 13$. This simplifies down to $y = -5(x+2)^2 + 33$. 
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