Quartic equation with two obvious real solutions x=-2 and x=1, Expand the quartic and divide by (x+2)(x-1) to obtain the quadratic equation x^2-x-3=0. This gives an additional pair of conjugate solutions by application of the formula for quadratic equations x=(1±√13)/2. If x=-2 and x=1 are not immediately obvious consider this: there is a 4th order term on the LHS and a linear term on the RHS. Therefore the absolute values of the solutions will be "small". If x are suspected to be integer: small squares (LHS) are 0 1 4 8. x+3=1 --> x=-2 and x+3=4 -->x-1 are the two obvious integer solutions.
General Solution: (x^2-a)^2-x-a=0 Expand: x^4-2ax^2-x+a(a-1)=0. Factor into product of 2 quadratics: (x^2-x-a)(x^2+x-(a-1))=0. Solve the quadratic equations using the well known formula. x^2-x-a=0 --> x=(1±√(1+4a))/2. x^2+x-a(a-1)=0 --> x=(-1±√(4a-3))/2. For a=3 this gives the four solutions shown in the clip.
Easy solution -: (x² - 3)² = x + 3 x⁴ - 6x² - x + 6 = 0 x( x³-1)-6(x²-1)=0 x(x-1)(x²+x+1)-6(x-1)(x+1)=0 (x-1)(x³+x²-5x-6)=0 now find the 1st root of this cubic equation by hit and trail taking value of x as 0,1,-1,2 now as -2 we got the 1st root so our equation becomes (x-1)(x+2)(x²-x-3) now find roots of this quadratic equation using (-b ± √ (b2 - 4ac) )/2a we get x= (1+√13)/2 , (1-√13)/2 and our other two roots 1 , -2 as our final answer
Quartic equation with two obvious real solutions x=-2 and x=1, Expand the quartic and divide by (x+2)(x-1) to obtain the quadratic equation x^2-x-3=0. This gives an additional pair of conjugate solutions by application of the formula for quadratic equations x=(1±√13)/2.
If x=-2 and x=1 are not immediately obvious consider this: there is a 4th order term on the LHS and a linear term on the RHS. Therefore the absolute values of the solutions will be "small". If x are suspected to be integer: small squares (LHS) are 0 1 4 8. x+3=1 --> x=-2 and x+3=4 -->x-1 are the two obvious integer solutions.
Very nice! ❤
General Solution: (x^2-a)^2-x-a=0
Expand: x^4-2ax^2-x+a(a-1)=0. Factor into product of 2 quadratics:
(x^2-x-a)(x^2+x-(a-1))=0. Solve the quadratic equations using the well known formula.
x^2-x-a=0 --> x=(1±√(1+4a))/2.
x^2+x-a(a-1)=0 --> x=(-1±√(4a-3))/2.
For a=3 this gives the four solutions shown in the clip.
Very nice trick! ❤
Thank you teacher ❤
You're welcome 😊
(x² - 3)² = x + 3
x⁴ - 6x² + 9 = x + 3
x⁴ - 6x² - x + 6 = 0 ← it would be interesting to have 2 squares on the left side (because power 4 and power 2)
Let's rewrite the equation by introducing a variable "λ" which, if carefully chosen, will produce 2 squares on the left side.
Let's tinker a bit with x⁴ as the beginning of a square: x⁴ = (x² + λ)² - 2λx² - λ²
x⁴ - 6x² - x + 6 = 0 → where: x⁴ = (x² + λ)² - 2λx² - λ²
(x² + λ)² - 2λx² - λ² - 6x² - x + 6 = 0
(x² + λ)² - [2λx² + λ² + 6x² + x - 6] = 0 → let’s try to get a second member as a square
(x² + λ)² - [2λx² + 6x² + x + λ² - 6] = 0
(x² + λ)² - [x².(2λ + 6) + x + (λ² - 6)] = 0 → a square into […] means that Δ = 0 → let"s calculate Δ
Δ = (1)² - 4.[(2λ + 6).(λ² - 6)] → then, Δ = 0
1 - 4.[(2λ + 6).(λ² - 6)] = 0
4.[(2λ + 6).(λ² - 6)] = 1
8.[(λ + 3).(λ² - 6)] = 1
(λ + 3).(λ² - 6) = 1/8
λ³ - 6λ + 3λ² - 18 = 1/8
λ³ - 6λ + 3λ² - 18 - (1/8) = 0
λ³ - 6λ + 3λ² - (145/8) = 0
λ = - 5/2
Restart:
(x² + λ)² - [x².(2λ + 6) + x + (λ² - 6)] = 0 → where: λ = 1/2
[x² - (5/2)]² - [x².(2.{- 5/2} + 6) + x + ({- 5/2}² - 6)] = 0
[x² - (51/2)]² - [x².(- 5 + 6) + x + ({25/4} - {24/4})] = 0
[x² - (5/2)]² - [x² + x + (1/4)] = 0 ← we can recognize a square
[x² - (5/2)]² - [x + (1/2)]² = 0 → a² - b² = (a + b).(a - b)
[x² - (5/2) + x + (1/2)].[x² - (5/2) - x - (1/2)] = 0
(x² + x - 2).(x² - x - 3) = 0
First case: (x² + x - 2) = 0
x² + x - 2 = 0
Δ = 1 - (4 * - 2) = 1 + 8 = 9
x = (- 1 ± 3)/2
→ x = 1
→ x = - 2
Second case: (x² - x - 3) = 0
x² - x - 3 = 0
Δ = (- 1)² - (4 * - 3) = 1 + 12 = 13
x = (1 ± √13)/2
→ x = (1 + √13)/2
→ x = (1 - √13)/2
(X^2-3)^2-4 = X+3-4=X-1 . (X^2-1)(X^2-5) = (X-1). (X-1)[(X+1)(X^2-5) -1]=0 . X=1
Very nice trick! I really appreciate that ❤
@@SALogics Đi làm giờ mới về. Chúc thầy giáo khỏe.
Let y = x^2-3
then y^2 = x+3
hence y^2+y = x^2+x
(x-y)(x+y+1) = 0
(x-x^2+3)(x+x^2-2) = 0
(x^2-x-3)(x+2)(x-1) = 0
roots are
(1+√13)/2 , (1-√13)/2 , -2 ,1
Very nice trick! ❤
Easy solution -:
(x² - 3)² = x + 3
x⁴ - 6x² - x + 6 = 0
x( x³-1)-6(x²-1)=0
x(x-1)(x²+x+1)-6(x-1)(x+1)=0
(x-1)(x³+x²-5x-6)=0
now find the 1st root of this cubic equation by hit and trail taking value of x as 0,1,-1,2 now as -2 we got the 1st root so our equation becomes
(x-1)(x+2)(x²-x-3)
now find roots of this quadratic equation using (-b ± √ (b2 - 4ac) )/2a
we get x= (1+√13)/2 , (1-√13)/2 and our other two roots 1 , -2 as our final answer
Very nice trick! ❤
X=1
Very nice! ❤