You can
put this solution on YOUR website!x^4-5x^2+4=0
Let y=x^2
Then we get:
y^2-5y+4=0
(y-4)(y-1)=0
Therefore y=4,1
Therefore x^2=4 or 1
Since squares of negative numbers are also positive,we get:
x=+2,-2,+1,-1
Now out of these,-1 and -2 are both less than zero,ie.non-real
Hence real solutions for x will be:
x=1,2
Hope this helps,
Prabhat
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