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X^3+512=0 real or imaginary solution by factoring
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i am going to guess that this is the difference of two cubes
x^3 - 512 = 0 (x - 8)(x^2 + 8x + 64) = 0 x = 8 and x = (use quadratic roots formula: x = -b +/- √b^2 - 4ac all over 2a) = -8 +/- √(64 - 256) all over 2 = -8 +/- √-192 all over 2 = -8 +/- 6i√3 all over 2 i means imaginary number because you're taking the square root of a negative number (which is impossible without i) = -4 +/- 3i√3
No, it was just the problem given and told to find the real or imaginary solutions by factoring
How would you do x^4-3x^2=-2x^2
\[(x - 8)(x^2 + 8x + 64) = 0\] \(x-8=0\implies x=8\) or \(x^2+8x+64=0\) complete the square or use the quadratic formula
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