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x⁴ - 16 = x² - 4(x² + 4)(x² - 4) - (x² - 4) = 0(x² - 4)(x² + 3) = 0x² - 4 = 0 => *x = ± 2*x² + 3 = 0 => *x = ± i√3*
what about checking x3 and x4 (plus/minus i times square root 3?
I solve it in my head in just 5Seconds...😂😂😂😂.
x=+-2; i×(+-sqrt3)
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x⁴ - 16 = x² - 4
(x² + 4)(x² - 4) - (x² - 4) = 0
(x² - 4)(x² + 3) = 0
x² - 4 = 0 => *x = ± 2*
x² + 3 = 0 => *x = ± i√3*
what about checking x3 and x4 (plus/minus i times square root 3?
I solve it in my head in just 5Seconds...😂😂😂😂.
x=+-2; i×(+-sqrt3)