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determine the natur of the roots
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(2x+8)^2 =0 x= -4 and x=-4
first one divide by 4 and get \[x^2+8x+16=0\] then if you want the "nature of the zeros" compute \[b^2-4ac=64-64=0\]so there is one solution. meaning it is a perfect square namely \[(x+4)^2\] and the solution \[x=-4\]
second one \[3x^2+12x-3=0\] \[b^2-4ac=12^2-4\times 3\times -3=144+36=180\] so there are two real solutions
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so solve write \[x^2+4x-1=0\] \[x^2+4x=1\] \[(x+2)^2=1+2^2=1+4=5\] \[x+2=\pm\sqrt{5}\] \[x=-2\pm\sqrt{5}\]
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