sac8513 wrote:

Im not sure why I dont see this...I saw this in an answer explanation

x^2 > x breaks down to x < -1 or x > 1

i can break this apart to be x(x - 1) > 0 So x > 0 or x > 1 . Now theoretically i understand that any negative number squared, or to an even power, is positive but i dont see how x < -1 is derived from that inequality

help is appreciated. thanks

Hi,

Whenever we get a quadratic equation (or any polynomial for that matter) in an inequality, we calculate the roots merely to find out at which points on the x-axis (considering a polynomial in 'x') the equation will be equal to 0. These points at which at which the equation is 0, help us to identify the different intervals over which we have to check the consistency of the equation. Confusing? Let me explain it with the help of the example you have provided.

Considering the quadratic equation : \(x^2 - x > 0\)

First we will have to find the roots of this equation which you have rightly calculated to be 0 and 1.

Now on the x-axis, plot these two values of 'x' for which the equation becomes 0. This would divide the x-axis into three intervals. Namely :

1) x < 0

2) 0 < x < 1

3) x > 1

Now we have to check the consistency of each of these intervals by picking any value contained in the interval and seeing whether it satisfies our original inequality : \(x^2 - x > 0\)

Interval 1 : x < 0

Lets pick x = - 1

\((-1)^2 - (-1) > 0\) ---> \(2 > 0\)

Since it satisfies our inequality, x must be true for all x < 0

Interval 2 : 0 < x < 1

Lets pick x = 0.5

\((0.5)^2 - 0.5 > 0\) ---> \(-0.25 > 0\)

Since this is false, x cannot be true for all 0 < x < 1

Interval 3 : x > 1

Lets pick 2

\((2)^2 - 2 > 0\) ---> \(2 > 0\)

Since it satisfies our inequality, x must be true for all x > 1

Thus the range of x such that x satisfies the inequality \(x^2 - x > 0\) is :

1) x < 0

2) x > 1This is the method for solving any inequality dealing with polynomials.

Note : The equation is satisfied for all negative values of x and not just those less than -1 as you have mentioned in your post.

A quicker method that I use to find out the range of 'x' in quadratic equations is thinking of the equation to be a parabola of the form : \(y = ax^2 + bx + c\)

Now there are a few things to note here :

1) When 'a' is positive, the parabola will open upwards.

2) When 'a' is negative, the parabola will open downwards (although you can multiply the entire inequality by -1 and make 'a' positive, resulting in a parabola that faces upwards. The end result will be the same).

3) The intersection of the parabola with the x axis will represent the two roots of the equation.

Attachment:

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Again considering the example you provided : \(y = x^2 - x\) is our parabola and our inequality becomes \(y > 0\). Thus we have to find the regions for which the value of 'y' is positive.

Now since the coefficient of \(x^2\) is positive, the parabola will 'open upwards'. Also, the parabola will intersect the x-axis at points 0 and 1.

Attachment:

Untitled1.png [ 8.64 KiB | Viewed 1263 times ]
Since we want to find the regions for which 'y' is positive and we know that it is an upward facing parabola, it is obvious that the regions we need to consider are

x < 0 and

x > 1 because in the region

0 < x < 1, y will be negative.

This is a much quicker method than sitting and checking all the intervals individually.

_________________

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