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x² + [3x/(x - 3)]² = 16
x² + [(3x)²/(x - 3)²] = 16
x² + [9x²/(x - 3)²] = 16
[x².(x - 3)² + 9x²] / (x - 3)² = 16
x².(x - 3)² + 9x² = 16.(x - 3)²
x².(x - 3)² - 16.(x - 3)² + 9x² = 0
(x - 3)².(x² - 16) + 9x² = 0
(x² - 6x + 9).(x² - 16) + 9x² = 0
x⁴ - 16x² - 6x³ + 96x + 9x² - 144 + 9x² = 0
x⁴ - 6x³ + 2x² + 96x - 144 = 0 → the aim, if we are to continue effectively, is to eliminate terms to the 3rd power.
x⁴ - 6x³ + 2x² + 96x - 144 = 0 → let: x = z - (b/4a) → where:
b is the coefficient for x³, in our case: - 6
a is the coefficient for x⁴, in our case: 1
x⁴ - 6x³ + 2x² + 96x - 144 = 0 → let: x = z - (- 6/4) → x = z + (3/2)
[z + (3/2)]⁴ - 6.[z + (3/2)]³ + 2.[z + (3/2)]² + 96.[z + (3/2)] - 144 = 0
[z + (3/2)]².[z + (3/2)]² - 6.[z + (3/2)]².[z + (3/2)] + 2.[z² + 3z + (9/4)] + 96z + 144 - 144 = 0
[z² + 3z + (9/4)].[z² + 3z + (9/4)] - 6.[z² + 3z + (9/4)].[z + (3/2)] + 2z² + 6z + (9/2) + 96z + 144 - 144 = 0
[z² + 3z + (9/4)].[z² + 3z + (9/4)] - 6.[z² + 3z + (9/4)].[z + (3/2)] + 2z² + 6z + (9/2) + 96z + 144 - 144 = 0
[z⁴ + 3z³ + (9/4).z² + 3z³ + 9z² + (27/4).z + (9/4).z² + (27/4).z + (81/16)] - 6.[z³ + (3/2).z² + 3z² + (9/2).z + (9/4).z + (27/8)] + 2z² + 6z + (9/2) + 96z + 144 - 144 = 0
[z⁴ + 6z³ + (27/2).z² + (27/2).z + (81/16)] - 6.[z³ + (9/2).z² + (27/4).z + (27/8)] + 2z² + 102z + (9/2) = 0
z⁴ + 6z³ + (27/2).z² + (27/2).z + (81/16) - 6z³ - 27z² - (81/2).z - (81/4) + 2z² + 102z + (9/2) = 0
z⁴ - (23/2).z² + 75z - (171/16) = 0 ← no itel to the 3rd power → it would be interesting to have 2 perfect squares on the left side
Let's rewrite the equation by introducing a variable "λ" which, if carefully chosen, will produce 2 squares on the left side
z⁴ = (z² + λ)² - 2λz² - λ² → we are going to rewrite the equation
z⁴ - (23/2).z² + 75z - (171/16) = 0 → where: z⁴ = z² + λ)² - 2λz² - λ²
(z² + λ)² - 2λz² - λ² - (23/2).z² + 75z - (171/16) = 0
(z² + λ)² - [2λz² + λ² + (23/2).z² - 75z + (171/16)] = 0
(z² + λ)² - [z².{2λ + (23/2)} - 75z + {λ² + (171/16)}] = 0 → if the contain of the second brackets is a perfect square Δ = 0
Let's calculate Δ
Δ = (- 75)² - 4.{ [2λ + (23/2)] * [λ² + (171/16)] } → then Δ = 0
(- 75)² - 4.{ [2λ + (23/2)] * [λ² + (171/16)] } = 0
4.{ [2λ + (23/2)] * [λ² + (171/16)] } = (- 75)²
[2λ + (23/2)] * [λ² + (171/16)] = (- 75)²/4
[2λ + (23/2)] * [λ² + (171/16)] = 5625/4
2.[λ + (23/4)] * [λ² + (171/16)] = 5625/4
[λ + (23/4)] * [λ² + (171/16)] = 5625/8
λ³ + (171/16).λ + (23/4).λ² + (3933/64) = 5625/8
λ³ + (23/4).λ² + (171/16).λ - (41067/64) = 0
(64/64).λ³ + (368/64).λ² + (684/64).λ - (41067/64) = 0
64λ³ + 368λ² + 684λ - 41067 = 0
→ λ = 27/4
Recall from the equation
(z² + λ)² - [{2λ + (23/2)}.z² - 75z + λ² + (171/16)] = 0 → where: λ = 27/4 to get a perfect square
[z² + (27/4)]² - [{(27/2) + (23/2)}.z² - 75z + (27/4)² + (171/16)] = 0
[z² + (27/4)]² - [(50/2).z² - 75z + (729/16) + (171/16)] = 0
[z² + (27/4)]² - [(50/2).z² - 75z + (900/16)] = 0
[z² + (27/4)]² - [25z² - 75z + (30/4)²] = 0
[z² + (27/4)]² - [25z² - 75z + (225/4)] = 0
[z² + (27/4)]² - 25.[z² - 3z + (9/4)] = 0
[z² + (27/4)]² - 25.[z² - 3z + (3/2)²] = 0
[z² + (27/4)]² - 5².[z - (3/2)]² = 0 ← now, there are 2 squares → a² - b² = (a + b).(a - b)
{ [z² + (27/4)] + 5.[z - (3/2)] }.{ [z² + (27/4)] - 5.[z - (3/2)] } = 0
[z² + (27/4) + 5z - (15/2)].[z² + (27/4) - 5z + (15/2)] = 0
[z² + 5z - (3/4)].[z² - 5z + (57/4)] = 0
First case: [z² + 5z - (3/4)] = 0
z² + 5z - (3/4) = 0
Δ = (5)² - 4.[1 * - (3/4)] = 25 + 3 = 28
z = (- 5 ± √28)/2
z = (- 5 ± 2√7)/2 → recall: x = z + (3/2)
x = (- 5 ± 2√7)/2 + (3/2)
x = (- 52 ± 2√7 + 3)/2
→ x = - 1 ± √7
Second case: [z² - 5z + (57/4)] = 0
z² - 5z + (57/4) = 0
Δ = (- 5)² - 4.[1 * (57/4)] = 25 - 57 = - 32 = 32i²
z = (5 ± i√32)/2
z = (5 ± 4i√2)/2
Recall: x = z + (3/2)
First case:
z = (- 5 ± 2√7)/2 → recall: x = z + (3/2)
x = (- 5 ± 2√7)/2 + (3/2)
x = (- 5 ± 2√7 + 3)/2
x = (- 2 ± 2√7)/2
→ x = - 1 ± √7
Second case:
z = (5 ± 4i√2)/2 → recall: x = z + (3/2)
x = (5 ± 4i√2)/2 + (3/2)
x = (5 ± 4i√2 + 3)/2
x = (8 ± 4i√2)/2
→ x = 4 ± 2i√2
Wow! Great job, thank you so much.
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Sir x=4+_√2í hai......
Apny Galat likha ha
Not at all!
x = 4 ± 2√2i is the correct answer
Because
We have x = 8 ± √16x2 i / 2
x = 8 ± 4√2i / 2
x = 2(4 ± 2√2i) / 2
x = 4 ± 2√2i
I hope you understood
Thanks for asking
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x ≠ 3, a = 3x/(x-3) => ax-3a = 3x => ax = 3(a+x) , x²+a² = 16, p = ax, s = a+x =>
p = 3s , s² = x²+a²+2ax = 16+2p => s²=16+6s => s² -6s -16 = 0 = (s-8)(s+2) =>
s = a+x = 8, p = ax = 3s = 24 => (8-x)x = 24 => x²-8x+24 = 0 = (x-4)²-16+24 > 0 =>
s = a+x = -2, p = ax = 3s = -6 => (-2-x)x = -6 => x²+2x-6 = 0 = (x+1)²-7 => x = -1±V7
Wow! thank you
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Love your videos. They remind me how much I forgot 😂
Thank you so much!
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What is the meaning of this equation?
Which equation?
I have solved this equation to get a quadratic equation.
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❤❤❤
thanks