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Let t=(5+26–√−−−−−−−√)x⟹(5−26–√−−−−−−−√)x=1tt=(5+26)x⟹(5−26)x=1t
Thus the equation becomes,
t+1t=10⟹t2−10t+1=0t+1t=10⟹t2−10t+1=0 which is a quadratic equation and have roots t=5+26–√,5−26–√t=5+26,5−26
If t=5+26–√⟹(5+26–√−−−−−−−√)x=5+26–√⟹(5+26–√)1−x/2=1⟹1−x/2=0⟹x=2t=5+26⟹(5+26)x=5+26⟹(5+26)1−x/2=1⟹1−x/2=0⟹x=2
If t=5−26–√⟹(5+26–√−−−−−−−√)x=5−26–√=15+26√⟹(5+26–√)1+x/2=1⟹x=−2t=5−26⟹(5+26)x=5−26=15+26⟹(5+26)1+x/2=1⟹x=−2
Thus, x=2,−2x=2,−2 are the two solutions.
Verification:
Putting x=2x=2 in the original equation gives L.H.S=5+26–√+5−26–√=10=5+26+5−26=10=R.H.S
down vote
accepted
Let t=(5+26–√−−−−−−−√)x⟹(5−26–√−−−−−−−√)x=1tt=(5+26)x⟹(5−26)x=1t
Thus the equation becomes,
t+1t=10⟹t2−10t+1=0t+1t=10⟹t2−10t+1=0 which is a quadratic equation and have roots t=5+26–√,5−26–√t=5+26,5−26
If t=5+26–√⟹(5+26–√−−−−−−−√)x=5+26–√⟹(5+26–√)1−x/2=1⟹1−x/2=0⟹x=2t=5+26⟹(5+26)x=5+26⟹(5+26)1−x/2=1⟹1−x/2=0⟹x=2
If t=5−26–√⟹(5+26–√−−−−−−−√)x=5−26–√=15+26√⟹(5+26–√)1+x/2=1⟹x=−2t=5−26⟹(5+26)x=5−26=15+26⟹(5+26)1+x/2=1⟹x=−2
Thus, x=2,−2x=2,−2 are the two solutions.
Verification:
Putting x=2x=2 in the original equation gives L.H.S=5+26–√+5−26–√=10=5+26+5−26=10=R.H.S
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