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

1 Answer
Oct 10, 2016

Answer:

#y=-1 or 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+SX+P=0#
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if we can find two real numbers #a 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+3y+2# has
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#S=3 and P=2#
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So,
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#color(blue)(y^2+3y+2=(y+2)(y+1))#

second: solve

#y^2+3y+2=0#
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#(y+2)(y+1)=0#
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#y+2=0rArry=-2#
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Or
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#y+1=0rArry=-1#