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Completing the square to find the zeros 3. a^2+2a-3=0

User Walshy
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Let's solve the equation by completing the square:


\begin{gathered} a^2+2a-3=0 \\ a^2+2a=3 \\ a^2+2a+((2)/(2))^2=3+((2)/(2))^2 \\ a^2+2a+1^2=3+1^2 \\ (a+1)^2=4 \\ \sqrt[]{(a+1)^2}=\sqrt[]{4} \\ \lvert a+1\rvert=2 \\ a+1=\pm2 \\ \text{then} \\ a+1=2 \\ a=2-1 \\ a=1 \\ or \\ a+1=-2 \\ a=-2-1 \\ a=-3 \end{gathered}

therefore the solutions are a=1 and a=-3

User Yuriy Zaletskyy
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