## anonymous 5 years ago can you help me to solve the following inequality: x^2 >=16

1. anonymous

Can you show the work, please?

2. anonymous

graph $y=x^2-16$ and you will see a parabola that faces up and crosses the x-axis at -4 and 4. from the picture you will see that it is positive if x<-4 or if x> 4

3. anonymous

analitically

4. anonymous

there is another more complicated way to do this, but if you know what the graph of $y=x^2-16$ looks like it is simple.

5. anonymous

well step number one is going to be to write $x^2-16\geq0$

6. anonymous

so you are going to have to work with that in any case. you cannot, for example, take the square root of both sides.

7. anonymous

I need to do it analitically

8. anonymous

then you can factor $x^2-16=(x+4)(x-4)\geq 0$

9. anonymous

got it

10. anonymous

Im such a fool

11. anonymous

then you can look at the sign of each factor. yes?

12. anonymous

you are still going to get positive, negative, positive on $(-\infty,-4), (-4,4), (4,\infty)$ respectively

13. anonymous

thanks

14. anonymous

welcome

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