SOLUTION: hello, can you help me to solve this question? x^2+9>0 thank you

Question 126327This question is from textbook
: hello, can you help me to solve this question?
x^2+9>0
thank you This question is from textbook

Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
A little "tricky". The answer to this question is that any real value of x (positive or
negative) will satisfy this equation. Why? Because x^2 will always be a positive number. And
when you add 9 to it, the result is still a positive number. And a positive number is greater
than zero.
.
Let's just try a few values of x to demonstrate this.
.
Suppose x is +10. Then x^2 + 9 = 10^2 + 9 = 100 + 9 = +109. This certainly is greater than
zero as is to be the case required by the problem.
.
Suppose x is +1. Then x^2 + 9 = 1^2 + 9 = 1 + 9 = +10. This is still greater than zero.
.
Suppose x equals zero. Then x^2 + 9 = 0^2 + 9 = 0 + 9 = +9. Still greater than zero.
.
Now suppose x = -1. Then x^2 + 9 = (-1)^2 + 9 = 1 + 9 = +10. Still greater than zero.
.
Then suppose x = -10. Then x^2 + 9 = (-10)^2 + 9 = 100 + 9 = +109. And it is also greater
than zero.
.
In fact if we graph y = x^2 + 9 the graph looks like:
.

.
You can see from this graph that x^2 + 9 never has a value less than 9 no matter what the
value along the x axis is. Therefore, x^2 + 9 is always greater than zero.
.
Hope this helps you to understand the problem a little better.
.
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