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cebroski
 one year ago
Best ResponseYou've already chosen the best response.2It is not true that x<a

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

cebroski
 one year ago
Best ResponseYou've already chosen the best response.2However, perhaps you meant to say x < a?

liliegirl
 one year 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... :))

vishal011998
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.1Take the square root of both sides.

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

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

triciaal
 one year 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

RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.1Consider 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

RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.1Since 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

RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.1Now, just like solving linear inequality, we have x < a, *or* x>a as the solution.

RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.1Note 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.

RolyPoly
 one year ago
Best ResponseYou've already chosen the best response.1So, 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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