Asked by Anonymous
the sides of a triangle have lengths 4x+1, 2x+1 and 6x-1. if the lengths of the longest side is 6x-1, what values of x make the triangle obtuse?
Answers
Answered by
Reiny
(6x-1)^2 > (4x+1)^2 + (2x+1)^2 ,
36x^2 - 12x + 1 = 16x^2 + 8x + 1 + 4x^2 + 4x + 1
6x^2 - 24x -1 > 0
suppose we solve the corresponding equation
6x^2 - 24x -1 = 0
x = (24 ± √600)/12
=(24 ± 10√6)/24
= (12 ± 5√6)/12 = appr 2.0206 or a negative
but clearly 6x-1 > 0
x > 1/6
so it is obtuse for x > 2.0206
36x^2 - 12x + 1 = 16x^2 + 8x + 1 + 4x^2 + 4x + 1
6x^2 - 24x -1 > 0
suppose we solve the corresponding equation
6x^2 - 24x -1 = 0
x = (24 ± √600)/12
=(24 ± 10√6)/24
= (12 ± 5√6)/12 = appr 2.0206 or a negative
but clearly 6x-1 > 0
x > 1/6
so it is obtuse for x > 2.0206
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