SOLUTION: {{{ x^2+3x+2=0 }}}.

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Question 579742: +x%5E2%2B3x%2B2=0+.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
+x%5E2%2B3x%2B2=0+
+x%5E2+%2B+3x+=+-2+
Complete the square:
Take 1/2 of the coefficient of x,
square it, and add it to both sides
+x%5E2+%2B+3x+%2B+%283%2F2%29%5E2+=+-2+%2B+%283%2F2%29%5E2+
+x%5E2+%2B+3x+%2B+9%2F4+=+-%288%2F4%29+%2B+9%2F4+
+%28+x+%2B+3%2F2+%29%5E2+=+%281%2F2%29%5E2+
Take the square root of both sides
+x+%2B+3%2F2+=+1%2F2+
+x+=+-%283%2F2%29+%2B+1%2F2+
(1) +x+=+-1+ ( one root of the equation )
and, taking the negative square root,
+x+%2B+3%2F2+=+-%281%2F2%29+
+x+=+-%283%2F2%29+-+1%2F2+
(2) +x+=+-2+ ( the other root of th equation )
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Rewriting (1) and (2),
(1) +x+%2B+1+=+0+
(2) +x+%2B+2+=+0+
and
+%28+x+%2B+1+%29%2A%28+x+%2B+2+%29+=+x%5E2+%2B+3x+%2B+2+