Understanding quadratic inequalities: 2 questions about solution for 2x^2+1 > 4x

[Alex McCaile]

Hi!
This is my first ever question on freeMATHhelp. I have been working with ineqialities examples, and I struggle a little bit with the logic. I will first write the text og the example and what I struggle with.

The solution text: " Given inequality 2x^2+1 > 4x The inequality 2x^2+1 > 4x is equivalent to 2x^2-4x+1>0. The corresponding quadric equation 2x^2-4x+1=0, which is of the form Ax^2 + Bx + C = 0, can be solved by the quadric formula. The given inequality can be expressed in the form (x-(1+1/2sqrt(2))(x-(1-1/2sqrt(2)) > 0. This is satisfied if both factors on the left side are positive or if both are negative. Therefore, we require that either x<1-1/2sqrt(2) or x>1+1/2sqrt(2). The solution set is the union of intervals ( -∞, 1-1/2sqrt(2)) ∪ (1+1/2*sqrt(2), ∞)"

This is what I dont understand:

1. "This is satisfied if both factors on the left side are positive or if both are negative". Why should the factors both be positive or negative?
2. What is the union of intervals? What is the logic behind unions?

Sorry if I made grammar mistakes, English is not my first language.

 

[stapel]

The solution text: " Given inequality 2x^2+1 > 4x The inequality 2x^2+1 > 4x is equivalent to 2x^2-4x+1>0. The corresponding quadric equation 2x^2-4x+1=0...

Does "quadric" mean "quadratic" (that is, of degree two)?

...which is of the form Ax^2 + Bx + C = 0, can be solved by the quadric formula.

I'm fairly certain that they are referring to the Quadratic Formula.

The given inequality can be expressed in the form (x-(1+1/2sqrt(2))(x-(1-1/2sqrt(2)) > 0.

If you plug the quadratic into the Quadratic Formula, you will get the following two solutions:

[imath]\qquad x = \dfrac{-(-4) \pm \sqrt{(-4)^2\;}}{2(2)}[/imath]

[imath]\qquad \qquad = \dfrac{4 \pm \sqrt{8\;}}{4}[/imath]

[imath]\qquad \qquad = 1 \pm \dfrac{\sqrt{2\;}}{2}[/imath]

Just as a solution of [imath]x = 3[/imath] equates to a factor of [imath]x - (3)[/imath], so also the two solutions spit out by the Quadratic Formula can be turned into factors of the original quadratic.

This is satisfied if both factors on the left side are positive or if both are negative.

The product of opposite-signed numbers yields a "minus" number; the only way for a product to yield a "plus" number is for the two factors to have the same sign.

Therefore, we require that either x<1-1/2sqrt(2) or x>1+1/2sqrt(2).

The quadratic is required to be greater than zero. The quadratic has a positive leading coefficient, so it graphs as an upward-opening parabola. The only place upward-opening parabolas are above the [imath]x[/imath]-axis is on the two sides of the parabola. In the middle, between the zeroes, the parabola is below the axis.

Plot the two solutions on the number line. Graph the parabola. Where is the parabola positive?

The solution set is the union of intervals ( -∞, 1-1/2sqrt(2)) ∪ (1+1/2*sqrt(2), ∞)"

What is the union of intervals? What is the logic behind unions?

The union of two intervals is the set of all values which are in one of the intervals or in the other of the intervals. Since the solution set for this quadratic inequality is the [imath]x[/imath]-values where the parabola is above the axis, then the solution set is all [imath]x[/imath]-values in the interval to the left or else in the interval to the right.

No [imath]x[/imath]-value can be in both of these separate intervals at the same time, so set-intersection is not correct for this solution set.

 

[Steven G]

+*+ = +
+*- = -
-*+ = -
-*- = +

 

[blamocur]

What is the union of intervals? What is the logic behind unions?

Can you see which area(s) correspond to positive values of your quadratic function? Can you see that it does not fit in a single interval?

 

[Alex McCaile]

Does "quadric" mean "quadratic" (that is, of degree two)?



I'm fairly certain that they are referring to the Quadratic Formula.



If you plug the quadratic into the Quadratic Formula, you will get the following two solutions:

[imath]\qquad x = \dfrac{-(-4) \pm \sqrt{(-4)^2\;}}{2(2)}[/imath]

[imath]\qquad \qquad = \dfrac{4 \pm \sqrt{8\;}}{4}[/imath]

[imath]\qquad \qquad = 1 \pm \dfrac{\sqrt{2\;}}{2}[/imath]

Just as a solution of [imath]x = 3[/imath] equates to a factor of [imath]x - (3)[/imath], so also the two solutions spit out by the Quadratic Formula can be turned into factors of the original quadratic.



The product of opposite-signed numbers yields a "minus" number; the only way for a product to yield a "plus" number is for the two factors to have the same sign.



The quadratic is required to be greater than zero. The quadratic has a positive leading coefficient, so it graphs as an upward-opening parabola. The only place upward-opening parabolas are above the [imath]x[/imath]-axis is on the two sides of the parabola. In the middle, between the zeroes, the parabola is below the axis.

Plot the two solutions on the number line. Graph the parabola. Where is the parabola positive?



The union of two intervals is the set of all values which are in one of the intervals or in the other of the intervals. Since the solution set for this quadratic inequality is the [imath]x[/imath]-values where the parabola is above the axis, then the solution set is all [imath]x[/imath]-values in the interval to the left or else in the interval to the right.

No [imath]x[/imath]-value can be in both of these separate intervals at the same time, so set-intersection is not correct for this solution set.

Hi! Sorry, I made a typo, yes, I actually meant quadratic.

Does "quadric" mean "quadratic" (that is, of degree two)?



I'm fairly certain that they are referring to the Quadratic Formula.



If you plug the quadratic into the Quadratic Formula, you will get the following two solutions:

[imath]\qquad x = \dfrac{-(-4) \pm \sqrt{(-4)^2\;}}{2(2)}[/imath]

[imath]\qquad \qquad = \dfrac{4 \pm \sqrt{8\;}}{4}[/imath]

[imath]\qquad \qquad = 1 \pm \dfrac{\sqrt{2\;}}{2}[/imath]

Just as a solution of [imath]x = 3[/imath] equates to a factor of [imath]x - (3)[/imath], so also the two solutions spit out by the Quadratic Formula can be turned into factors of the original quadratic.



The product of opposite-signed numbers yields a "minus" number; the only way for a product to yield a "plus" number is for the two factors to have the same sign.



The quadratic is required to be greater than zero. The quadratic has a positive leading coefficient, so it graphs as an upward-opening parabola. The only place upward-opening parabolas are above the [imath]x[/imath]-axis is on the two sides of the parabola. In the middle, between the zeroes, the parabola is below the axis.

Plot the two solutions on the number line. Graph the parabola. Where is the parabola positive?



The union of two intervals is the set of all values which are in one of the intervals or in the other of the intervals. Since the solution set for this quadratic inequality is the [imath]x[/imath]-values where the parabola is above the axis, then the solution set is all [imath]x[/imath]-values in the interval to the left or else in the interval to the right.

No [imath]x[/imath]-value can be in both of these separate intervals at the same time, so set-intersection is not correct for this solution set.

Hi! Sorry, I made a typo, yes, I meant quadratic. I just needed a clarification of the solution, and your answer definitely helped me. Thank you very much!