Question

Determine the number of real root of the equation :
(x^2+1)^2 - x^2 = 0

Answer 1

Answer:

Step-by-step explanation: (x^2+1)^2 - x^2 = 0.

Let f(x) = (x^2+1)^2 - x^2  = $x^4+x^2+1$ = 0

That is

f(x) = $x^4+x^2+1$

Replace x with + x and -x respectively

f(x) = $x^4+x^2+1$

f(-x) = $(-x)^4+(-x)^2+1=x^4+x^2+1$

Since, there is no sign change in f(x) and f(-x), so by Descartes rule, there is no real roots.

Hence, the number of real roots in (x^2+1)^2 - x^2 = 0 is 0.

Answer 2

Answer:

-½

Step-by-step explanation:

(x²+1)² - x² =0

= x²+ 2x + 1 - x² = 0

by cancelling +x² and -x² ,

=> 2x + 1 = 0

=> 2x = -1

=> x = -½

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