Question

21^√x+20^√x=29^√x. solve for x​

Answer 1

Given : 21^√x+20^√x=29^√x

To Find:  Solve for x :

Solution:

21^√x+20^√x=29^√x

Pythagorean triplet :

21²  + 20²  = 29²

Comparing both

√x  = 2

Squaring both sides

=> x = 2²

=> x = 4

Value of x = 4

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