Question

Please solve this problem. Show that $\sf{(x^2+1)^2-x^2}$ is positive for reals.

Answer 1

Step-by-step explanation:

(x^2+1)^2-x^2}[

=f(x^2-y^2)

= your answer is 0

Answer 2

Concept to be used here:

• Factorization.
• Inequality.
• Quadratic Functions.

Solving the Problem

By Factorization Method

$\sf{(x^2+x+1)(x^2-x+1)}$

Because of

• the negative discriminant

• positive highest coefficient

we know that

$\sf{x^2+x+1>0}$

$\sf{x^2-x+1>0}$

The Conclusion

After multiplying together

$\sf{(x^2+x+1)(x^2-x+1)>0}$

In another way.

$\sf{(x^2+1)^2-x^2=x^4+x^2+1}$

• A number squared is always positive or 0.

We have

$\sf{x^4\geq 0}$

$\sf{x^2\geq 0}$

Adding together

$\sf{x^4+x^2}\geq 0$

So we get

$\sf{x^4+x^2+1\geq 1>0}$