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Question

How many real roots will be there for the equation x10+x8+x6+x4+x2+8=0? ___

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Solution

f(x)=x10+x8+x6+x4+x2+8=0
No. of sign changes = 0.
Hence, the number of positive real roots =0.
Now, f(x)=x10+x8+x6+x4+x2+8=0
No. of sign changes = 0.
Hence, the number of negative real roots =0.
The number of real roots for the given equation is 0.


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