Question 338745
2x^4-16x=0
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Factor:
2x(x^3-8) = 0
x(x^3-8) = 0
x(x-2)(x^2+2x+4) = 0
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x = 0, or x = 2 or x = the solutions for the quadratic:
x = [-2+-sqrt(4-4*4)]/2
x = [-2+-sqrt(-12)]/2
x = [-2+-2isqrt(3)]/2
x = [-1+-isqrt(3)]
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Cheers,
Stan H.