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cebroski Group TitleBest ResponseYou've already chosen the best response.2
It is not true that x<a
 11 months ago

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

cebroski Group TitleBest ResponseYou've already chosen the best response.2
However, perhaps you meant to say x < a?
 11 months ago

liliegirl Group TitleBest ResponseYou've already chosen the best response.0
You can prove that using absolute value... That's what I learn from our analytic geometry class... :))
 11 months ago

vishal011998 Group TitleBest ResponseYou've already chosen the best response.0
Yeah thanx@cebroski but i asked as it was used by friend to solve an answer...
 11 months ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
Take the square root of both sides.
 11 months ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
sqrt(x^2) = x
 11 months ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
Hence x > a Take cases
 11 months ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
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<a
 11 months ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.1
There really is no statement that sums up all three except x > a
 11 months ago

triciaal Group TitleBest ResponseYou've already chosen the best response.0
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
 11 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
Consider 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
 11 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
Since 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
 11 months ago

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

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
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.
 11 months ago

RolyPoly Group TitleBest ResponseYou've already chosen the best response.1
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.
 11 months ago
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