You can find two real solution by observing that, if you set x=0, the equation becomes (-4)(-5)(-6)(-7) i.e. 840, which is 1680/2 From this, any number which doubles the left side of the equation is a solution, then x=-1 and x=12 are the first two roots. By the magnitude of the numbers you may infer that the remaining two solutions will not be real numbers, but complex ones. To see that, we may use the real roots obtained to run Ruffini's rule on the polynomial expression you obtain after multiplying (x-4)(x-5)(x-6)(x-7)=1680 (x^2-9x+20)(x^2-13x+42)=1680 x^4-22x^3+179x^2-638x+840=1680 x^4-22x^3+179x^2-638x-840=0 Running twice the Ruffini's rule we obtain at the end (x+1)(x-12)(x^2-11x+70)=0 In this, we can see that the 2nd degree polynomial expression has two complex solutions (1/2)(11+13i) and (1/2)(11-13i)
To summarise your first stage a little, for integer x, we are seeking 4 consecutive integers whose product is 1680. As you indicate, we quickly find that 5*6*7*8=1680, leading to x=12 or x=-1.
Substitute x=y+5.5 into the given equation and rearrange to (y^2-1/4)(y^2-9/4)-1680=y^4-(5/2)y^2-26871/16=0 y^2=(5/2±164/2)/2=169/4 or -159/4, y=±13/2 or ±i√159/2, and x=(11±13)/2=12 or -1 or x=(11±i√159)/2.
The solution clue lies on the consecutive four factors. Factors of 1680 are 5 × 6 × 7 × × 8 =( x - 4) (x - 5) (x -6) (x -7 ) (12 -7) (12 -6) (12 - 5) (12-4) Implies x= 12 (don't go for wastage of time )
There is a simpler solution in 5 basic steps: (x-4) (x-5) (x-6) (x-7) = 1680 Key insight: 4 left terms are symmetric around 5.5 But let's scale first to avoid factions 1) Multiply both sides by 2^4 = 16 (absorb the 2's into each left term) (2x-8) (2x-10) (2x-12) (2x-14) = 16×1680 2) Define substitution y = 2x - 11 to reveal symmetry: (y+3) (y+1) (y-1) (y-3) = 16×1680 (y²-9) (y²-1) = 16×1680 3) Define another substitution z=y²-1 z² - 8 z = 16×1680 4) Simply solve quadratic equation z = 4 ± SQ(16×1680 + 16) z = 4 ± 4 SQ(1681) = 4 ± 4 × 41 5) z must be positive real number thus z = 168 => y = 13 => x = 12
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I guess complex answers are beyond the scope of this video.
Thanks for sharing.
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You can find two real solution by observing that, if you set x=0, the equation becomes (-4)(-5)(-6)(-7) i.e. 840, which is 1680/2
From this, any number which doubles the left side of the equation is a solution, then x=-1 and x=12 are the first two roots. By the magnitude of the numbers you may infer that the remaining two solutions will not be real numbers, but complex ones. To see that, we may use the real roots obtained to run Ruffini's rule on the polynomial expression you obtain after multiplying
(x-4)(x-5)(x-6)(x-7)=1680
(x^2-9x+20)(x^2-13x+42)=1680
x^4-22x^3+179x^2-638x+840=1680
x^4-22x^3+179x^2-638x-840=0
Running twice the Ruffini's rule we obtain at the end
(x+1)(x-12)(x^2-11x+70)=0
In this, we can see that the 2nd degree polynomial expression has two complex solutions
(1/2)(11+13i) and (1/2)(11-13i)
To summarise your first stage a little, for integer x, we are seeking 4 consecutive integers whose product is 1680. As you indicate, we quickly find that 5*6*7*8=1680, leading to x=12 or x=-1.
Substitute x=y+5.5 into the given equation and rearrange to (y^2-1/4)(y^2-9/4)-1680=y^4-(5/2)y^2-26871/16=0
y^2=(5/2±164/2)/2=169/4 or -159/4, y=±13/2 or ±i√159/2, and x=(11±13)/2=12 or -1 or x=(11±i√159)/2.
The solution clue lies on the consecutive four factors.
Factors of 1680 are 5 × 6 × 7 × × 8 =( x - 4) (x - 5) (x -6) (x -7 )
(12 -7) (12 -6) (12 - 5) (12-4)
Implies x= 12 (don't go for wastage of time )
Thank you for your solution
You're very welcome! 🙏❤️🙏
(x-4)*(x-7)•(x-5)•(x-6)=1680
(X²-11x+28)•(x²-11x+30)=1680
X²-11x+28=t
t•(t+2)=1680
t²+2t-1680=0
(t+1)²-1681=0
(t+1-41)(t+1+41)=0
t=40
t=-42
X²-11x+28=40
X²-11x+28=-42
X²-11x-12=0
X²-11x-70=0
Shu tenglamani yechish yetadi.
Thank you so much
You're very welcome! 🙏❤️🙏
T cant be taken as common it should be t×t+2 some thing is wrong
X equal.....
12
There is a simpler solution in 5 basic steps:
(x-4) (x-5) (x-6) (x-7) = 1680
Key insight: 4 left terms are symmetric around 5.5
But let's scale first to avoid factions
1) Multiply both sides by 2^4 = 16 (absorb the 2's into each left term)
(2x-8) (2x-10) (2x-12) (2x-14) = 16×1680
2) Define substitution y = 2x - 11 to reveal symmetry:
(y+3) (y+1) (y-1) (y-3) = 16×1680
(y²-9) (y²-1) = 16×1680
3) Define another substitution z=y²-1
z² - 8 z = 16×1680
4) Simply solve quadratic equation
z = 4 ± SQ(16×1680 + 16)
z = 4 ± 4 SQ(1681) = 4 ± 4 × 41
5) z must be positive real number thus
z = 168 => y = 13 => x = 12
You forgot the option of y=-13, which leads to x =-1, the second real solution - as found in the original video.
Maranatha JESUS
ОК.