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if x²>a² then x>a and x<-a how?

Mathematics
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It is not true that x<-a
For example, if x=10 and a=2, then 10² > 2² because 100 > 4, but 10 < -2 is not true.
However, perhaps you meant to say -x < -a?

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Other answers:

You can prove that using absolute value... That's what I learn from our analytic geometry class... :))
Yeah thanx@cebroski but i asked as it was used by friend to solve an answer...
Take the square root of both sides.
sqrt(x^2) = |x|
Hence |x| > |a| Take cases
If x>0 and a>0 then x>a If x<0 and a>0 then -x>a If x<0 and a<0 then x
There really is no statement that sums up all three except |x| > |a|
x^2 is positive and (-a)^2 is positive x is less than -a so you a squaring a "bigger negative number" for ex -5 is <-2 (-5)^2 >(-2)^2 5 is bigger than 2
Consider x²>a² x² - a² > 0 (x-a)(x+a) > 0 So, we know the zeros are a and -a. Now, draw a graph for x² - a² = 0 (please forgive my poor drawing) |dw:1389442252543:dw|
Since x² - a² > 0, we can shade the regions which are bounded by x² - a² = 0 and x-axis with *POSITIVE* y. We get: |dw:1389442338672:dw|
Now, just like solving linear inequality, we have x < -a, *or* x>a as the solution.
Note that since we are NOT looking for the overlapping part as the solution, we are NOT using "and" when we write the solution, i.e. "x>a and x<-a". Instead, we need to use *OR* when we write the solution, i.e. x>a *or* x<-a.
So, in @cebroski 's example, even though x>-a is NOT true when x=10 and a=2, it is TRUE that x>a when x=10 and a=2. Therefore, the solution still holds for the quadratic inequality.

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