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

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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) About Me  (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.
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Let's just try a few values of x to demonstrate this.
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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.
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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.
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Now suppose x = -1. Then x^2 + 9 = (-1)^2 + 9 = 1 + 9 = +10. Still greater than zero.
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Then suppose x = -10. Then x^2 + 9 = (-10)^2 + 9 = 100 + 9 = +109. And it is also greater
than zero.
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In fact if we graph y = x^2 + 9 the graph looks like:
.
graph%28500%2C500%2C+-15%2C15%2C-5%2C120%2Cx%5E2+%2B+9%29
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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.
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Hope this helps you to understand the problem a little better.
.