kartik222 wrote:
Hi Bunuel,
What exactly is the solution for:
x^2<x
From what I understand, it is:
x^2-x<0 => x(x-1)<0
either x<0 , x-1>0 => x>1 ..OR.. x>0 , x-1<0 => x<1, so there can be two possibilities here.
BUT saw somewhere that x^2<x ===> 0<x<1, is this true? according to me this is just one possibility.
In the similar way can u please also explain x^2>x (signed changed)
I really appreciate your help.
-K
For any quadratic inequation, here are steps that can be followed to solve it:
1) Write the inequation in the form
x^2+bx+c>0 or
x^2+bx+c<02) Find the roots of the quadratic equation
x^2+bx+c=0.3) Sketch the graph of the function
y=x^2+bx+c, which is an upward parabola, intersecting the X-axis at the values found at step 2) above.
4) Read the solution from the graph.
In the beginning, you should practice the steps and draw the graph. Later, you can just imagine it and automatically write down the solution.
Let's apply the above steps to the inequation
x^2<x.1)
x^2-x<02)
x^2-x=0 or
x(x-1)=0; the two roots are 0 and 1.
3) For the graph, see the attached drawing.
4) From the graph, we can see that
x^2-x is negative (the graph is below the X-axis) if
x is between the roots 0 and 1.
Therefore, the solution to the given inequation is
0<x<1.If instead of
< there was
\leq, the solution would be
0\leq{x}\leq{1}, meaning we should include the values of
x for which the expression becomes 0. These are exactly the roots of the quadratic equation we solved at step 2).
If we have to solve the inequation
x^2-x>0, we look again at the graph of the parabola and find those values of
x for which the graph is above the Y-axis.
We obtain the solution
x<0 OR
x>1.Every upward parabola has its two "arms" above the X-axis for values of
x outside the interval defined by the two roots, and it has the "arc" below the X-axis for values of
x between the two roots.
This is the "justification" for how the sign of a quadratic expression changes in the three intervals defined by the two roots of the quadratic. Once you understand this, it will be really easy to quickly write down the solution for any similar inequation, and you will need just to imagine the parabola.
Attachments
Parabola-signs.jpg [ 13.11 KiB | Viewed 275 times ]
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