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A = 4B
C = B-20
solve for B
pythagorean formula is C^2 = A^2 + B^2
substituting to get all unknowns in terms of B, equation becomes
(B-20)^2 = (4B)^2 + B^2
multiplying out, formula becomes
B^2 - 40B + 400 = 4B^2 + B^2
combining like terms, formula becomes
-4B^2 - 40B + 400 = 0
multiplying left hand side of equation and right hand side of equation by -1 and formula becomes
4B^2 + 40B - 400 = 0
dividing both sides of the equation by 4 and the equation becomes
B^2 + 10B - 400 = 0
answer is not evident by inspection so use quadratic formula to get the roots.
quadratic formula is
and
where
-b = -10
b^2-4*a*c = (10^2 - 4*1*(-100) = 100 - (-400) = 500
2a = 2
solving we get B = 6.180339888 or B = -16.180339888.
since B can't be negative, the answer is B = 6.180339888.
substituting for B in the original equation derived from the pythagorean formula, we get
B^2 + 10*B - 100 = (6.180339888)^2 + 10*(6.180339888) - 100
= 38.19660113 + 61.80339888 - 100 = 0 which becomes
100 - 100 = 0
which becomes 0 = 0 proving formula is correct.
answer is B = 6.180339888 = 6.18 rounded to nearest hundredth.
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