Quadratic Inequallities

[Anatanoshi]

Recently i learned about Quadratic Inequallities and i can't understand when i should flip the > or < sign.

Can anyone help me with that?

[Sriman Dutta]

A quadratic inequality is an inequation of the form :

[math] ax^2 + bx + c > 0 (< 0, >=0, <=0) [/math]

The first step is to solve the quadratic equation [math]ax^2+bx+c=0[/math] and get the roots [math]r_1[/math] and [math]r_2[/math].

Then put it as:

[math](x-r_1)(x-r_2) > 0[/math]

Now solve it using the method of intervals.

Remember that we can solve such inequalities with method of intervals only when the roots are real. Thus, discriminant should be >= 0.

Edited May 15, 2017 by Sriman Dutta

[Country Boy]

The direction of the inequality is 'flipped' if and only if each side is multiplied or divided by a negative number. The difficulty, of course, is that if you multiplied or divided by an expression in "x", whether the expression is positive or negative depends upon x. So basically, you need to determine when each such expression is positive or negative.

 

For example, to solve [math]x^2- 5x+ 6= (x- 5)(x- 6)> 0[/math], I can see that, if x< 5, x- 5< 0 so dividing both sides by it will "flip" the direction: x- 6< 0 which gives x< 6. Of course, if x< 5, then we automatically have x< 6 so one solution is x< 5. If x> 5, then x-5 is positive and dividing by that does not flip the direction: x- 6> 0 which gives x> 6. Of course, if x>6 then we must have x> 5 so x>> 6 is a solution. The solution set is {x| x< 5}U{x|x> 6}.

 

Another way: we can argue that the product of two numbers is positive if and only if the two numbers have the same sign- both positive or both negative. That is, either x-5> 0 and x- 6> 0 or x-5< 0 or x-6< 0. The first pair give x>5 and x> 6 which are both true for x> 6. The second pair give x< 5 and x< 6 which are both true for x< 5. Again, the solution set is {x| x< 5}U{x> 6}.

 

Finally, we could use Sriman Duttas method. First solve (x- 5)(x- 6)= 0. The solution set is {5, 6}. Since the function is continuous, those are the only points at which the function can change from negative to positive or from positive to negative. That is, there must be a single sign on each of the intervals [math](-\infty, 5)[/math], [math](5, 6)[/math], and [math](6, \infty)[/math]. It is sufficient to choose one point in each interval to determine what that sign is. x= 0 is in [math](-\infty, 5)[/math] and (0- 6)(0- 5)= 30> 0 so (x- 5)(x- 6) is positive for all x< 5. x= 5.5 is in [math](5, 6)[/math] and (5.5- 6)(5.5- 5)= (-0.5)(0.5)= -0.25< 0 so (x- 5)(x- 6) is negative for all x between 5 and 6. Finally, 7 is in [math](6, \infty)[/math] and (7- 5)(7- 6)= 2> 0 so (x- 5)(x- 6) is positive for all x> 6. Once again, the solution set is {x| x< 5}U{x> 6}.

Edited May 15, 2017 by Country Boy

[Anatanoshi]

I think i got it.

Thank you!

[KipIngram]

For example, to solve [math]x^2- 5x+ 6= (x- 5)(x- 6)> 0[/math], I can see that, if x< 5, x- 5< 0 so dividing both sides by it will "flip" the direction: x- 6< 0 which gives x< 6. Of course, if x< 5, then we automatically have x< 6 so one solution is x< 5. If x> 5, then x-5 is positive and dividing by that does not flip the direction: x- 6> 0 which gives x> 6. Of course, if x>6 then we must have x> 5 so x>> 6 is a solution. The solution set is {x| x< 5}U{x|x> 6}.

 

Um, isn't x^2-5x+6 = (x-2)(x-3)?