draw the graph of and and find for what range of x does . I think this method is much easier.
find the solution set and draw the solution in the real number line.
|x+2| < |x (x - 1)|
all i can think is to transpose the right side of the in inequality like
|x+2| - |x (x - 1)|< 0 equate by zero
then i am stuck here... if i get the value of x by quadratic ... i don't kn0w to do this as a simple solving.
thanks
First, notice that is positive if x is negative,
but is positive if
so you have a bound for which you can remove the modulus symbols.
Then both are positive also if
Both are never negative for the same x, since makes the right side positive.
But you can write for
However the right is negative for
so then you can write
And a fourth way! First solve the equation |x+ 2|= |x(x-1)|. If x+ 2 and x(x- 1) have the same sign, then that equation becomes x+ 2= x(x- 1)= x^2- x or x^2- 2x= 2. Completing the square, x^2- 2x+ 1= (x- 1)^2= 3 so x= 1+sqrt(3) or x= 1- sqrt(3). If x+ 2 and x(x- 1) have opposite signs, then the equation becomes -(x+ 2)= x(x- 1) or -x- 2= x^2- x so that x^2= -2 which is never true.
Now the point is that the inequality can change from ">" to "<" and vice-versa only where we have "=" and that is at x= 1- sqrt(3) and 1+ sqrt(3). -1 < 1- \sqrt(3) and if x= -1 the inequality becomes |-1+ 2|< |(-1)(-1-1)| or "1< 2" which is true. That means that the inequality is true for all x< 1- \sqrt{3}.
Now check a single value of x between 1- sqrt(3) and 1+ sqrt(3) to determine whether or not all points in that interval satisfy the inequality and then check a single value of x larger than 1+ sqrt(3).
Yes jam2011,
good method, a variation of Plato's.
However, you've mistyped it on the 2nd line...
therefore
Find the roots and the inequality is true for x to the left of the smaller root
and to the right of the larger root.