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Help with math


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Step Step:

The projection of the tangent length on the Y axis is X ^ 2 + 9.

The derivative is 2X, and so the tangent length projection on the X axis is X ^ 2 + 9) / 2X).

The sum of these two factors squared gives us the function we are trying to minimize (the root can already be thrown at this point, and also say that we are only looking for the positive X due to the symmetry). Note that after uploading the square of these expressions as they get the maximum holding of X is 4 and the minimum is 2-. After the derivation, the 3 and 3 holdings are obtained, hence the sixth hold of the final polynomial.

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I got the equation


⁶ + 73t⁴ - 81 = 0

Its solutions:





i * sqrt8 / 3

i * sqrt8 / 3-

You have the 1, -1 solutions in the equation, which is the solution, but the other solutions are different (- + sqrt (0.125) i). Someone here is wrong. DCAG, please list your solution. My path is the same as Vic's.

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The tangent equation: L = 2ax-a ^ 2 + 9

Launch Point: (a, a ^ 2 + 9), v. The tangent of the tangent with the X axis: ((a ^ 2-9) / 2a, 0)

The distance equation between them: y = sqrt [((a ^ 2-9) / 2a-a) ^ 2 + (0- (a ^ 2 + 9)) ^ 2]

After arrangement, it looks like this: y = (a ^ 2 + 9) * sqrt (1 + 4a ^ 2) / 2a

tea. a! = 0

הנגזרת: y'=[2a*2a-2(a^2+9)]*sqrt(1+4a^2)/4a^2 + (a^2+9)*8a/(2a*2*sqrt(1+4a^2)) = (a^2-9)*sqrt(1+4a^2)/2a^2 + (2a^2+18)/sqrt(1+4a^2)

We will compare y '= 0 and after arranging we will come to the equation: 8a ^ 4 + a ^ 2-9 = 0

נגדירa^2=t ונמצא: t1=1,t2=-18/16

I did not refer to the composite solution but you can see that the results of a are:


So many signs ..., just that they don't jump parentheses from start to finish = /.

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