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Find the value of x. AD = 2x + 4 DB = 6x - 20 http://image.tutorvista.com/content/feed/tvcs/diagonalrectangle45.JPG
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nope it was just that...
\[(6x-20)(6x-20)=(2x-4)(2x-4)+AB^2\]\[36x^2-120x-120x+400=4x^2-8x-8x+16+AB^2\]\[(36x^2-4x^2)+(-240x+16x)+(400-16)=AB^2\] \[32x^2-224x+384=AB^2\] \[x^2-7x+12=AB^2\] \[(x-3)(x-4)=AB^2\] |dw:1405431728494:dw| |dw:1405431805619:dw|
that's the only thing given sorry o-o
@Anime_Basics i made a mistake i wrote (2x-4) instead of writing (2x+4) (2x+4)(2x+4)+AB^2=36x^2-240x+400 4x^2+8x+8x+16+AB^2=36x^2-240x+400 32x^2-256x+384=AB^2 x^2-8x+12=AB^2 sqrt(x^2-8x+1)=AB
(x-2)(x-6) and square root 6 and 2
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