# How do you solve y^2+3y+2=0?

Oct 10, 2016

$y = - 1 \mathmr{and} y = - 2$

#### Explanation:

To solve this equation first,we should factorize it.
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In this exercise we can't factorize using common factor nor applying
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Polynomial identities , so I will apply trial and error method instead.
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First: Factorize
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Trial and error method :
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${X}^{2} + S X + P = 0$
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if we can find two real numbers $a \mathmr{and} b$ whose sum is S and
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product is P then color(blue)(X^2+SX+P=0=(X+a)(X+b)
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In the given equation the polynomial
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${y}^{2} + 3 y + 2$ has
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$S = 3 \mathmr{and} P = 2$
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So,
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$\textcolor{b l u e}{{y}^{2} + 3 y + 2 = \left(y + 2\right) \left(y + 1\right)}$

second: solve

${y}^{2} + 3 y + 2 = 0$
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$\left(y + 2\right) \left(y + 1\right) = 0$
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$y + 2 = 0 \Rightarrow y = - 2$
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Or
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$y + 1 = 0 \Rightarrow y = - 1$