SOLUTION: What are the values of x for which 8-x^2 < 5? I tried solving for x and got x < sqrt(3). That is only one piece in this puzzle, though. How would I go about answering this que

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Question 629494: What are the values of x for which 8-x^2 < 5?
I tried solving for x and got x < sqrt(3). That is only one piece in this puzzle, though. How would I go about answering this question?
This problem comes from a practice test I took today in preparation for an upcoming math placement exam.
Thank you!

Answer by jsmallt9(3758) (Show Source):
You can put this solution on YOUR website!
Actually, your "one piece" is not right. I suspect you multiplied or divided both sides by a negative at some point and forgot the special rule that tells you to reverse the inequality.

There are several ways to figure this out. One of them would be...
Make one side of the inequality zero. The fast way to do this would be to subtract 5 from each side. But I prefer to avoid the special rule. So I like to make sure the variable term has a positive coefficient. So I am going to subtract 8 and add on both sides:

Now I'm going to use the difference of squares pattern, to factor the right side, You might wonder: "But 3 is not a perfect square!?" And you'd be right. It's not the square of a whole number. But every positive number is the square of something. In this case 3 is the square of . So we can factor the right side with this pattern:

The reasons we got a zero and then factored are that the inequality above tells us something important. It tells us that a product (the right side) is greater than zero. A synonym for "greater than zero" is positive. So this tells us that a product of two numbers is positive. You have probably known for a long time that to get a positive result when multiplying two numbers you must have:
Two positive numbers; or
Two negative numbers
So our solution to this problem will be "all the x values that make both factors positive or both factors negative." All we have to do is express this logic in the form a inequalities we can solve.

For "both factors are positive" I hope the following makes sense:
(x+sqrt(3) > 0 and x-sqrt(3)>0)
For "both factors are negative" I hope the following makes sense:
(x+sqrt(3) < 0 and x-sqrt(3)<0)
For "both factors positive or both factors negative" I hope that the following makes sense:
(x+sqrt(3) > 0 and x-sqrt(3)>0) or (x+sqrt(3) < 0 and x-sqrt(3)<0)
Now we just solve these:
(x > -sqrt(3) and x>sqrt(3)) or (x < -sqrt(3) and x
The first pair both say that x is greater than something. With the "and" between them we are interested only in x's that are greater than both. I hope it makes sense that if x is greater than the larger number then it would automatically be greater than the smaller number, too. So is all we need.

The second pair both say that x is less than something. With the "and" between them we are interested only in x's that are less than both. I hope it makes sense that if x is less than the smaller number then it would automatically be less than the larger number, too. So is all we need.

So our simplified solution is:
(x>sqrt(3)) or (x < -sqrt(3))