First search for integer solutions. Since none is found try factorization into two quadratic terms with integer coefficients, (x^2 + px + q)(x^2 - px + r) = x^4 + (q+r-p^2)x^2 - (q-r)px + qr = x^4 -12x -5. We must have qr = -5 with q+r positive (equal to p^2). The only integer solution is q=5, r=-1, which gives p^2=4. This matches the condition that (q-r)p=12 when p=2. Hence, (x^2 + 2x +5)(x^2 -2x-1) = x^4 -12x -5.
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First search for integer solutions. Since none is found try factorization into two quadratic terms with integer coefficients,
(x^2 + px + q)(x^2 - px + r) =
x^4 + (q+r-p^2)x^2 - (q-r)px + qr =
x^4 -12x -5.
We must have qr = -5 with q+r positive (equal to p^2). The only integer solution is q=5, r=-1, which gives p^2=4. This matches the condition that (q-r)p=12 when p=2. Hence,
(x^2 + 2x +5)(x^2 -2x-1) = x^4 -12x -5.
0:59 Let X^4 - 12X - 5 =(X^2 +aX+ p)(X^2 +bX + q) so
a+b=0 ; pq= -5;
p + q + ab = 0; aq + pb = - 12
solving a = 2; b = - 2 ; p = 5; q = - 1 ...
Beautiful exercise and very well resolved, thank you.
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x⁴ - 12x - 5 = 0 ← it would be interesting to have 2 squares on the left side (because power 4)
Let's rewrite the equation by introducing a variable "λ" which, if carefully chosen, will produce a square on the left side
Let's tinker a bit with x⁴ as the beginning of a square: x⁴ = (x² + λ)² - 2λx² - λ²
x⁴ - 12x - 5 = 0 → where: x⁴ = (x² + λ)² - 2λx² - λ²
(x² + λ)² - 2λx² - λ² - 12x - 5 = 0
(x² + λ)² - [2λx² + λ² + 12x + 5] = 0 → let's try to get a second member as a square
(x² + λ)² - [(2λ).x² + 12x + (λ² + 5)] = 0 → a square into […] means that Δ = 0 → let's calculate Δ
Δ = 12² - 4.[2λ * (λ² + 5)] → then, Δ = 0
12² - 8λ.(λ² + 5) = 0
8λ.(λ² + 5) = 12²
λ.(λ² + 5) = 18 → we can observ an obvious solution
λ = 2
Restart
(x² + λ)² - [(2λ).x² + 12x + (λ² + 5)] = 0 → where: λ = 2
(x² + 2)² - [4x² + 12x + 9] = 0 ← we can see a square inside […]
(x² + 2)² - (2x + 3)² = 0 → a² - b² = (a + b).(a - b)
[(x² + 2) + (2x + 3)].[(x² + 2) - (2x + 3)] = 0
(x² + 2 + 2x + 3).(x² + 2 - 2x - 3) = 0
(x² + 2x + 5).(x² - 2x - 1) = 0
First case: (x² + 2x + 5) = 0
x² + 2x + 5 = 0
Δ = 2² - (4 * 5) = 4 - 20 = - 16 = 16i²
x = (- 2 ± 4i)/2
→ x = - 1 ± 2i
Second case: (x² - 2x - 1) = 0
x² - 2x - 1 = 0
Δ = (- 2)² - (4 * - 1) = 4 + 4 = 8
x = (2 ± √8)/2
x = (2 ± 2√2)/2
→ x = 1 ± √2
Since there is no x^3, x^2 terms in the given equation,
x^4 - 12x - 5 = (x^2 + a/2)^2 - a(x + b/2)^2 = x^4 - abx + (a^2 - ab^2)/4
=> -ab = -12 & (a^2 - ab^2)/4 = -5 => a^2 - ab^2 = a^2 - a(12/a)^2 = -20
=> a^3 + 20a - 144 = (a - 4)(a^2 + 4a + 36) = 0 => a = 4 => b = 3
=> x^4 - 12x - 5 = (x^2 + 2)^2 - 4(x + 3/2)^2 = (x^2 + 2)^2 - (2x + 3)^2 = (x^2 + 2x + 5)(x^2 -2x - 1)
(x^4)^2 ➖( 12)^2 ➖ 5={x^16 ➖ 144} ➖ 5=128 ➖ (5)^2={128 ➖ 25}=103 10^10^3.2^5^2^5^3 1^1^2^1^3.23 (x ➖ 3x+2).
Taking super small steps and writing every single detail and repeating the writing of everything makes it boring and unnerving to the highest degree. I'd rather give up math altogether than see another RUclips clip by you. Not that you weren't correct. Just maddening...
When I move with speed avoiding trivial steps some viewers complain I'm too fast for them to understand . I think I need to strike a balance of all my viewership.