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anonymous
 2 years ago
if x²>a² then x>a and x<a how?
anonymous
 2 years ago
if x²>a² then x>a and x<a how?

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anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0It is not true that x<a

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0For example, if x=10 and a=2, then 10² > 2² because 100 > 4, but 10 < 2 is not true.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0However, perhaps you meant to say x < a?

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0You can prove that using absolute value... That's what I learn from our analytic geometry class... :))

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Yeah thanx@cebroski but i asked as it was used by friend to solve an answer...

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.1Take the square root of both sides.

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.1Hence x > a Take cases

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.1If 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<a

ParthKohli
 2 years ago
Best ResponseYou've already chosen the best response.1There really is no statement that sums up all three except x > a

triciaal
 2 years ago
Best ResponseYou've already chosen the best response.0x^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

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Consider x²>a² x²  a² > 0 (xa)(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

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Since x²  a² > 0, we can shade the regions which are bounded by x²  a² = 0 and xaxis with *POSITIVE* y. We get: dw:1389442338672:dw

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Now, just like solving linear inequality, we have x < a, *or* x>a as the solution.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0Note 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.

anonymous
 2 years ago
Best ResponseYou've already chosen the best response.0So, 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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