4x^2+8x+3 = 0
4x^2 + 8x + 3 + 1 = 0 + 1
4x^2 +8x + 4 = 1
(2x + 2)(2x + 2) = 1
(2x + 2)^2 = 1
(2x + 2) = Squareroot(1)
2x + 2 = 1
2x = - 1
x = -0.5
2007-04-30 03:31:20
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answer #1
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answered by OdinLowe 1
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to complete the square first move the 3 to the right side of the equation giving you
4x^2+8x = -3 divide everything by 4 to simplify term by x^2 giving you
x^2+2x= -3/4 now take the coefficent of the x term (2) and divide it by 2 then square it and add that to both sides of the equation... (2 divided by 2 is 1 and one squared is 1)
x^2+2x+1 = -3/4 +1 now the left side will be a perfect square...
(x+1)^2= 1/4... take the square root of both sides so you get two solutions
x+1 = 1/2 and x+1 = -1/2 solve both
x= -1/2 and x = -3/2
2007-04-30 10:42:27
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answer #2
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answered by sharkey 4
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4.(x² + 2x + 3/4) = 0
x² + 2x + 3/4 = 0
(x² + 2x + 1) - 1 + 3/4 = 0
(x + 1)² = 1/4
x + 1 = ± 1/2
x = - 1/2, x = - 3/2
2007-04-30 14:30:02
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answer #3
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answered by Como 7
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4x^2+8x+3=0
to complete the square means to express in the form of (ax+b)^2-b^2+3, which is equals (ax)^2+2axb+b^2-b^2+3
so you should get (2x)^2+2(2x)(2)+2^2-2^2+3, which is equals (2x)^2+2(2x)(2)+4-4+3=0, where ax=2x, and b=2
therefore, you will get:
(2x+2)^2-4+3=0
(2x+2)^2=1
2x+2=1 or 2x+2=-1
x= -1 x= -1.5
2007-04-30 10:35:52
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answer #4
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answered by elysium... 2
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