Ineqaulity x^2 <9

[raajpagare]

In the strategy guide no 3, there is an ineqaulity question

x^2<9

The solution of which after taking sqaure root of both sides is

|x| <3

But my question (which I know is wrong, but I cant find the flaw in the reasoning) is, shouldnt the result of taking the square root of both sides be

|x|<|3|

I know this is wrong, but why are we not using both the roots of 9 and only the positive one?

Thanks

[bacorcoran]

The solution does use both roots of 9! The absolute value in the left side of the inequality handles this.

Change it to x^2 = 9. Then the solution is x = 3 or -3. Or, more simply, |x| = 3.

Now go back to the inequality where the solution is |x| < 3. This solution can be described as all values on the number line between -3 and 3. This shows how both roots are considered in the solution.

Another way of addressing your question is to compare 3 vs. |3|. They are the same, so the absolute value in this case is redundant.

Your instincts are correct! You do need to consider both roots of 9! And the |x| is where this is covered!

[raajpagare]

Thanks for your reply

I am able to figure out x^2=9 problem this way

x^2=9
=> |x|=|3|
=> +-x=+-3
=>+x=+-3 , -x=+-3
=>x=3 or x=-3 , -x=3 i.e. x=-3 0r -x=-3 i.e x=3

So it boils down to
=>x= 3 or x=-3
So the equality does take both signs across the equality into account , and that makes sense.

But consider this in inequality


x^2<9
=> |x|<|3|
=> +-x<+-3
=>x<+-3 -x<+-3
=>x<3 or x<-3 -x<3 or -x<-3
=>x<3 0r x<-3 x>-3 or x>3
Where is the flaw in this approach.

Apologies for being a little dense :(
Thanks

[bacorcoran]

The mistake that you are making is saying that the square root of 9 is |3|. This is not correct. The square root of 9 is 3 or -3. And the way you consider both of these possible solutions is by using the absolute value on the variable, not both the variable and the real number.

|3| does not equal 3 or -3. It just equals 3.

Rather...if |x| = 3, then x can equal either 3 or -3.

For both the equation and the inequality, you've got to drop the |3|...that's your fundamental mistake. It's just 3.

[tim]

the other thing to keep in mind is that when you take all four combinations of +/- signs you have to make sure you change direction of the inequality where appropriate. this is what makes polynomial inequalities so difficult in general, and i recommend bacorcoran's approach: make it an equation and solve that. in this case, once you have determined that 3 and -3 are solutions to the equation, that splits the number line into three regions ( x<-3, -3<x<3, 3<x). just test one number from each of these regions so you can figure out which of them satisfy the conditions of the inequality..