Explain the positive values of a term in an inequality

Question

I'm solving this inequality:

$$x^4+5x^2 \geq 36$$

I have advanced to this point:

$$ x^4+5x^2-36\geq 0$$

I established that $$ u= x^2$$

Therefore $$ x^4+5x^2-36=(u-4)(u+9)$$ $$ (u-4)= (x+2)(x-2)$$ $$ (u+9)= (x^2+9)$$ $$(x+2)(x-2)(x^2+9)\geq 0$$ $$ x\leq -2 \ x\geq 2$$ As you can see I have already solved the inequality, but I need to explain why all the values of $(x^2+9)$ are positive without using square root.

Answer 1

Since

$$x^2 \ge 0,$$

we have

$$x^2+9 \ge 9 >0$$

Answer 2

Your inequality $$(x+2)(x-2)(x+9)\geq 0$$ Should have been $$(x+2)(x-2)(x^2+9)\geq 0$$

You can ignore $$(x^2+9)$$ because $$(x^2+9)\ge 9 >0$$it is always positive.

Solving $$(x+2)(x-2)\geq 0$$ is a piece of cake and I let you enjoy it.