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Find all values of t such that t - 1, t + 1, and 4 could be the lengths of the sides of a right triangle.

 Oct 22, 2017
 #1
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If "t" has to be an integer...it is impossible to have integer sides with "4" as a hypotenuse

 

So.......I will assme that either

 

4^2 + ( t - 1)^2  = ( t + 1)^2     or    4^2 + (t + 1)^2  = (t - 1)^2

 

But.......for the second to be true, t  = -4, which would give us negative side lengths

 

So.....we have

 

4^2 + ( t - 1)^2  = ( t + 1)^2

 

16 -2t  =  2t

 

16  = 4t   →   t  = 4

 

So....the other two sides are 3 and 5

 

If there are no restrictions on t, we have that

 

( t - 1)^2  +  ( t + 1)^2  = 4^2

 

2t^2  + 2  =  16

 

t^2 =  7   →   t =  √7

 

So.....the sides are   √7 - 1 , √7 + 1 ,  4

 

 

cool cool cool

 Oct 22, 2017
edited by CPhill  Oct 22, 2017

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