Maths help!!!?

Question

Do you know how to solve this?
|x^2 - x| < 2

Please explain step by step. Thank you.

Anyway, how do we know which is the right sign to use < or >?

Answer 1

|a| means +a or -a
similarly,|x^2-x| means +(x^2-x) or -(x^2-x)
case 1:considering +(x^2-x)
x^2-x<2
x^2-x-2<0
x^2-2x+x-2<0
x(x-2)+1(x-2)<0
(x-2)(x+1)<0
here,the product must be negative
therefore,either the first can be negative or the second can be negative
there fore the solution is x<2 or x<-1

case 2:considering -(x^2-x)
-x^2+x<2
x^2-x+2<0
x^2+2 but we know,x^2>x{except 1 but in this case because of +2,1 also does not satisfy the condition.}

there fore there are no values in case 2

there for the only solution of x is that it lies between -1 and 2{excluding both}

Answer 2

|x^2 - x| < 2
There are a few ways to do this question. I would recommend sketching the graph of y=|x^2 - x|. But I will show you another method as well.

Method 1: Sketch graph
Sketch the graph of y=|x^2 - x|
Then on the same axes, sketch the line y=2.
The line will intersect the curve at 2 points, 1 negative x value and 1 positive x value.
Do simultaneous eqns y=2 and y=x^2 - x or y= - (x^2 - x) to obtain the 2 x values, depending on which part of the curve the line intersects with.
Sounds confusing, but it will be clearer if I could draw it out for you. Sorry.

Method 2: By calculation considering 2 cases.
Case 1: x^2 - x >0
So,
|x^2 - x| < 2 is the same as x²-x<2 as x²-x is positive, so the modulus sign is not needed.

x²-x<2
x²-x-2<0
Sketch out y=x²-x-2 on a number line, the intersections with the x axis are x=2, x= - 1.

Alternate the + and minus signs, starting with + if the coefficient of the x ² term is positive. The 2 regions are -1 and 2 are obtained from the x intercepts of x²-x-2.
+ - +
-------|-------|----->
-1 2
Since you want to find x²-x-2<0, so you take the negative portion of the number line, which is -1
Case 2: x^2 - x <0
Since x^2 - x <0, so you need to add a negative sign infront of x^2 - x, to ensure that it will be positive (because negative negative = positive) as the modulus takes only the magnitude of the expression.

So, |x^2 - x| < 2 is the same as -(x²- x)< 2
-x²+x<2
x²-x+2>0
Complete the square, because x²-x+2 cannot be factorised completely.
x²-x+(1/2)²-(1/2)²+2>0
[x-(1/2)]²+7/3>0
As [x-(1/2)]²> or equals to 0 when x is any real number. Hence, the whole expression [x-(1/2)]²+7/3>0 for all real numbers.

Lastly, combine the solutions for the 2 cases, take the insection of the 2 solutions (meaning consider the AND scenario), hence your answer is -1
Hope this helps. =)

Answer 3

you must study the sign of x^2-x = x( x-1) which i

++++++0---------1+++++++++ (a)
so for x>=1 and <= 0
x^2-x-2<0 Find the roots of the first member
x=((1+-3)/2 x= 2 and x=-1 and the sign is

++++++++++ -1------------2+++++++++++
and is negative for -1 1 If 0 -x^2+x<2 so x^2+x+2 >0 which in the interval (b) is true as all summands are positive
so 0 -1 Looking at a former answer I would strongly remark that greater or smaller between complex numbers is NOT defined

Answer 4

The "<" sign means a conjunction, or "and".
If it was ">" instead then we would replace with OR.

|x(x-1)| < 2 -> x(x-1)<2 AND x(x-1)>-2

x(x-1)<2 -> x^2-x-2<0 -> -1
x(x-1)>-2 -> x^2-x+2>0 -> x is all reals

Therefore the solution is -1

Answer 5

-x^-2+x>2


I think this is inverse or inverted. Look it up in an Algebra textbook. It is like opposite numbers on a number line. You simply change the + to a - or vis versa.

Answer 6

x^2-x<+-2
x^2-x<2 or

x^2-x<-2

(x-2)(x+1)<0
thus one no is negative and other positive
so either
x-2<0 and x+1>0->x<2,x>-1
or
x-2>0 and x+1<0 ->x>2,x<-1

(x-0.5-1.322i)(x-0.5+1.322i)<0
so similarly to cases

x<0.5+1.322i,x>0.5-1.322i,
x>0.5+1.322i,x<0.5-1.322i