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Nice
Would have liked you to verify the complex answers as well. The question didn’t specify real numbers only.
Why did you used 12=8+4. This is not possible without to know that the answer is -2.
I new that as soon as looked at it.
2^6 2^2^3 1^2^1 2^1 (a ➖ 2a+1).
Minus 2?
Minus 2
x³ - x² + 12 = 0 Just try x = -/+ 1, 2, 3 and 4: x = 1: 1 - 1 + 12 = 12 fail x = 2: 8 - 4 + 12 = 0 failx = -2: -8 - 4 + 12 = 0 bingo ! So the first factor x1 = -2 1.x³ + -1.x² + 0.x + 12times -2 -2 6 -12 1 -3 6 0So (x + 2) . (x² - 3x + 6) = 0 Check: x³ - 3x² + 6x + 2x² - 6x + 12 = x³ - x² + 12 = 0 x² - 3x + 6 = 0 m + n = -3 ½(m + n) = - 3/2 and m . n = 6m . n = (-3/2 - d) . (-3/2 + d) = 6 so 9/4 - d² = 6 -> d² = 9/4 - 24/4 = -15/4 and d = iV15 / 2so x2 = 3/2 - iV15 / 2 and x3 = 3/2 + iV15 / 2
a³ - a² + 12 = 0a³ + 2³ - a² + 2² = 0(a³ + 2³) - (a² - 2²) = 0(a+2)(a²+2²-2a) - (a+2)(a-2) = 0(a+2)((a²+2²-2a) - (a-2)) = 0ok, a1 = -2, yeah ? 😊(a²+2²-2a) - (a-2) = 0a²+2²-2a - a + 2 = 0a² - 3a + 10 = 0....now comes the tedious part 😂D = b² - 4ac = - 31... those imaginary numbers, I can't be bothered with those...a2 = [3 + i*sqrt(31)] / 2a3 = [3 - i*sqrt(31)] / 2
...as it turned out I am too 99% - I failed... and failed where, here:a²+2²-2a - a + 2 = 0a² - 3a + 10 = 0I so very well did the hard bits but miserably failed at summing 2² + 2, instead of 6 I got 10...SHAME ON ME 😂
Nice
Would have liked you to verify the complex answers as well. The question didn’t specify real numbers only.
Why did you used 12=8+4. This is not possible without to know that the answer is -2.
I new that as soon as looked at it.
2^6 2^2^3 1^2^1 2^1 (a ➖ 2a+1).
Minus 2?
Minus 2
x³ - x² + 12 = 0
Just try x = -/+ 1, 2, 3 and 4:
x = 1: 1 - 1 + 12 = 12 fail
x = 2: 8 - 4 + 12 = 0 fail
x = -2: -8 - 4 + 12 = 0 bingo ! So the first factor x1 = -2
1.x³ + -1.x² + 0.x + 12
times -2 -2 6 -12
1 -3 6 0
So (x + 2) . (x² - 3x + 6) = 0 Check: x³ - 3x² + 6x + 2x² - 6x + 12 = x³ - x² + 12 = 0
x² - 3x + 6 = 0 m + n = -3 ½(m + n) = - 3/2 and m . n = 6
m . n = (-3/2 - d) . (-3/2 + d) = 6 so 9/4 - d² = 6 -> d² = 9/4 - 24/4 = -15/4 and d = iV15 / 2
so x2 = 3/2 - iV15 / 2 and x3 = 3/2 + iV15 / 2
a³ - a² + 12 = 0
a³ + 2³ - a² + 2² = 0
(a³ + 2³) - (a² - 2²) = 0
(a+2)(a²+2²-2a) - (a+2)(a-2) = 0
(a+2)((a²+2²-2a) - (a-2)) = 0
ok, a1 = -2, yeah ? 😊
(a²+2²-2a) - (a-2) = 0
a²+2²-2a - a + 2 = 0
a² - 3a + 10 = 0
....now comes the tedious part 😂
D = b² - 4ac = - 31... those imaginary numbers, I can't be bothered with those...
a2 = [3 + i*sqrt(31)] / 2
a3 = [3 - i*sqrt(31)] / 2
...as it turned out I am too 99% - I failed... and failed where, here:
a²+2²-2a - a + 2 = 0
a² - 3a + 10 = 0
I so very well did the hard bits but miserably failed at summing 2² + 2, instead of 6 I got 10...SHAME ON ME 😂