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Science & Mathematics by Anonymous 2018-07-05 20:06:14
Social Science
Find the complex numbers which satisfying the equation z^2-6z+36=0?
4 answers
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Anonymous
z² - 6z + 36 = 0 z² - 6z + 9 + 27 = 0 z² - 6z + 9 = - 27 (z - 3)² = - 27 (z - 3)² = 27i² (z - 3)² = (± i.√27)² z - 3 = ± i.√27 z = 3 ± i.√27 → you know that: √27 = √(9 * 3) = √9 * √3 = 3√3 z = 3 ± 3i.√3 z₁ = 3 + 3i.√3 z₂ = 3 - 3i.√3
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Anonymous
z² - 6z + 36 = 0 z² - 6z + 9 + 27 = 0 (z - 3)² - (3√3 i)² = 0 z = 3 + 3√3 i and z = 3 - 3√3 i
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Anonymous
z^2-6z=-36 z^2-6z+9=-27 (z-3)^2=-27 z = 3 ± 3i√3
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Anonymous
Roots (Zeroes) for the equation ax² + bx + c = 0 is x = [-b ± √(b² - 4ac)] / (2a) z² - 6z + 36 = 0 z = {-(-6) ± √[(-6)² - 4(1)(36)]} / [2(1)] z = [6 ± √(-108)] / 2 z = [6 ± 6(√3)i]/2 z = 3 + 3(√3)i or 3 - 3(√3)i