One of the few times I would sanction the use of the cubic equation. Just put 80x^3 -x +1 = 0 and solve. The answer you get is x= -0.25 meanwhile the other two roots are complex. Don’t bother with binomial expansion stuff which takes up a lot of time.
Hello, I am sorry but where did you pull that formula for the quadratic equation from???? First of all it is not 2C it is 2a. Second you are writing the equation for 1/x not X. The correct formula is X= 2a/[-b + - Sqrt(b*2 - 4ac)]. Replace 1/X by Y to minimize confusion from the beginning. Y*3 - Y*2 +80=0
Nice and elegant solution. I did also the Substitution like you. I factored to u^2 (1-u) = 80 = 16 * 5, so u = -4 (x = …) is a solution. Then I use polynomial Division by (u + 4) and Proof, that the polynomial reminder has no more roots.
One of the few times I would sanction the use of the cubic equation. Just put 80x^3 -x +1 = 0 and solve. The answer you get is x= -0.25 meanwhile the other two roots are complex. Don’t bother with binomial expansion stuff which takes up a lot of time.
Hello, I am sorry but where did you pull that formula for the quadratic equation from???? First of all it is not 2C it is 2a. Second you are writing the equation for 1/x not X. The correct formula is X= 2a/[-b + - Sqrt(b*2 - 4ac)]. Replace 1/X by Y to minimize confusion from the beginning. Y*3 - Y*2 +80=0
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(1/x)² - (1/x)³ = 80
(1/x)³ - (1/x²) + 80 = 0
1/x = u => x = 1/u
u³ - u² + 80 = 0
u³ + 64 - (u² - 16) = 0
(u + 4)(u² - 4u + 16) - (u + 4)(u - 4) = 0
(u + 4)(u² - 5u + 20) = 0
u = - 4 => *x = -1/4*
u² - 5u + 20 = 0 => complex roots
I solved it the same way.
Nice and elegant solution.
I did also the Substitution like you.
I factored to u^2 (1-u) = 80 = 16 * 5, so u = -4 (x = …) is a solution.
Then I use polynomial Division by (u + 4) and Proof, that the polynomial reminder has no more roots.
One look and I can see -1/4 is one of the roots.
(1/x)² - (1/x)³ = 80 → let: a = 1/x → where: x ≠ 0
a² - a³ = 80
a³ - a² + 80 = 0
a³ - (5a² - 4a²) + 80 = 0
a³ - 5a² + 4a² + 80 = 0
a³ - 5a² + 4a² + 80 + (20a - 20a) = 0
a³ - 5a² + 4a² + 80 + 20a - 20a = 0
a³ - 5a² + 20a + 4a² - 20a + 80 = 0
(a³ - 5a² + 20a) + (4a² - 20a + 80) = 0
a.(a² - 5a + 20) + 4.(a² - 5a + 20) = 0
(a + 4).(a² - 5a + 20) = 0
First case: (a + 4) = 0
a + 4 = 0
a = - 4 → recall: a = 1/x
→ x = - 1/4
Second case: (a² - 5a + 20) = 0
a² - 5a + 20 = 0
Δ = (- 5)² - (4 * 20) = 25 - 80 = - 55 = 55i²
a = (5 ± i√55)/2 → recall: a = 1/x
x = 2/(5 ± i√55)
First possibility: x = 2/(5 + i√55)
x = 2.(5 - i√55)/[(5 + i√55).(5 - i√55)]
x = 2.(5 - i√55)/[25 + 55]
x = 2.(5 - i√55)/80
→ x = (5 - i√55)/40
Second possibility: x = 2/(5 - i√55)
x = 2.(5 + i√55)/[(5 - i√55).(5 + i√55)]
x = 2.(5 + i√55)/[25 + 55]
x = 2.(5 + i√55)/80
→ x = (5 + i√55)/40
Awesome 👍😎💯
you made an error on the quadratic equation - it isn’t divided by 2c , it should be 2a
It is correct. Multipy the quadratic eqn. by x^2.Eqn. will be cx^2+bx+a=0.
Discriminant over 2a not 2c
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please redo this video - your videos are generally quite good however this one you failed to correct the error with the quadratic equation
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2c? what?
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You are pronouncing the letter B as the letter P!
Local dialect interference.