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Because the original equation has identical a^4 terms on both sides of the equation. Hence the problem is really only a cubic equation, hence the three solutions.
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Where do you get these algebra olympiads exercises from?
a^4 = (a - 1)^4 = a^4 - 4a^3 + 6a^2 - 4a + 1
4a^3 - 6a^2 + 4a - 1 = 0
(2a - 1)(2a^2 - 2a + 1) = 0
a = 1/2, (1 +/- i)/2
Can you just expand the right side using Pascal's Triangle.... Then work it from there?
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from which grade is this?
▫a²=±(a-1)²;
▫a²₁=(a₁-1)²=a²₁-2a₁+1;
▫1-2a₁=0;
▫2a₁=1;
▫a₁=½;
▫a²=-(a-1)²=-(a²-2a+1)=2a-a²-1;
▫2a²-2a+1=0;
▫D=4-4*2*1=4-8=-4;
▫√D=i√2;
▫a=(2±i√2)/4=2/4±i√2/4;
▫√2/4=√(2/16)=1/√8=1/(2√2);
▫a=½±i/(2√2);
▫a₂=½+i/(2√2);
▫a₃=½-i/(2√2).
I'm curious as to why only 3 solutions whereas we started with a^4
Because the original equation has identical a^4 terms on both sides of the equation. Hence the problem is really only a cubic equation, hence the three solutions.
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a=0 is also a solution
The substitution of zero into 'a' results in an incorrect equation.
0⁴ = (0 - 1)⁴
0 = (-1)⁴
0 = 1
1/2
trivial solution
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А = 0,5. Ich nicht see video
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