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
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.
I guess complex answers are beyond the scope of this video.
Thanks for sharing.
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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
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.
Thank you for your solution
T cant be taken as common it should be t×t+2 some thing is wrong
Maranatha JESUS