Finding the Range of x Values for Quadratic Inequalities

This is exactly the same as the question "for what values of x does (x-2)(x+1)<0?"II) x(x-2)>0. What are the values of x for which (x-2)(x+1)>0?III) 2-x>=3. What are the values of x for which (x-2)(x+1)>=3?IV) x(x-2)<3. What are the values of x for which (x-2)(x+1)<3?For II, III, and IV, you'll need to draw a number line and mark the places where the par
  • #1
greener1993
43
0
Ive got a few questions id like checking please but i start with the one I got no clue about :S

1)Slove the equation x(x-2)=2-x
So i asumme i slove it to zero, x^2-x-2=0
B)Use the solution to part A and the illustrated grapg to write down the solutions of
I) X(x-2)<2-x
II)X(x-2)>0
III)2-x>3 (greating or the same than)
Iv) x(x-2)<3
The picture is the attached image.

Ones i need checking are.

1) Slove theses simultaneous equations
X-y^2=10
X-2y=2

x=2+2y
(2-2y)+y^2=10 or 2-2y=10-y^2
Y^2+2y-8=0
(y+4)(y-2)
So y=-4 or y=2
for y=-4 x=2-2(-4), x= 10
For y= 2 x=2-2(2), x=-2, The problem here is that they both fit into the x=2+2y equation but not the other one, is this correct?


2) sloved an equation and got 4+-Root 20/2. How can this be simplyfied? i got it down to
2+-2root 2.5

Thank you for your help
 

Attachments

  • SAM_0454.JPG
    SAM_0454.JPG
    11.7 KB · Views: 431
Last edited:
Physics news on Phys.org
  • #2
Redo your simultaneous equations one - your + and - signs are incorrect almost from the beginning.

Not sure what #2 is, since you didn't put the original problem into your post.
 
  • #3
greener1993 said:
2) sloved an equation and got 4+-Root 20/2. How can this be simplyfied? i got it down to
2+-2root 2.5
You should learn LaTex. I don't know if you mean
[tex]\frac{4 \pm \sqrt{20}}{2}[/tex]
or
[tex]4 \pm \frac{\sqrt{20}}{2}[/tex]
or
[tex]4 \pm \sqrt{\frac{20}{2}}[/tex]
 
  • #4
greener1993 said:
Ive got a few questions id like checking please but i start with the one I got no clue about :S

1)Slove the equation x(x-2)=2-x
So i asumme i slove it to zero, x^2-x-2=0
The word is "solve", not "slove". While rewriting it as [itex]x^2- x- 2= 0[/itex] will work it might be simpler to notice that 2- x= -(x- 2) so the equation is x(x- 2)= -(x- 2). Now, either x= 2 or not. If [itex]x\ne 2[/itex], you can divide both sides by x- 2.

B)Use the solution to part A and the illustrated grapg to write down the solutions of
I) X(x-2)<2-x
II)X(x-2)>0
III)2-x>3 (greating or the same than)
"greater or the same as".
Iv) x(x-2)<3
The picture is the attached image.
The two points at which x(x- 2)= 2- x separate points where it is ">" from "<". Just check one value of x in each interval.

Ones i need checking are.

1) Slove theses simultaneous equations
X-y^2=10
X-2y=2

x=2+2y
(2-2y)+y^2=10 or 2-2y=10-y^2

You just said that x was 2+ 2y but then you replaced it in the second equation by 2- 2y!

Y^2+2y-8=0
(y+4)(y-2)
So y=-4 or y=2
for y=-4 x=2-2(-4), x= 10
For y= 2 x=2-2(2), x=-2, The problem here is that they both fit into the x=2+2y equation but not the other one, is this correct?
2) sloved an equation and got 4+-Root 20/2. How can this be simplyfied? i got it down to
2+-2root 2.5
[tex]\frac{4\pm\sqrt{20}}{2}= \frac{4\pm\sqrt{4(5)}}{2}[/tex][tex]= \frac{4\pm 2\sqrt{5}}{2}= 2\pm\sqrt{5}[/tex]
You cancel the "2" in front of the square root.
[tex]\frac{\sqrt{5}}{2}\ne \sqrt{\frac{5}{2}}[/tex]

Thank you for your help
 
  • #5
Your right i said x=2+2y then x=2-2y, but it would matter because I did, Y^2+2y-8=0 and work from that position with the correct symbol. Cheers for the help with the other one aswell.

However for the first question you didn't really help and I am still confused

P.S sorry about spelling i am dyslexic :(
 
  • #6
greener1993 said:
However for the first question you didn't really help and I am still confused

You do know the quadratic formula, don't you?

For equations of the form [itex]ax^2 + bx + c = 0[/tex], solve for x by;

[tex]x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]


So, in the case of [itex]x^2 - x -2 = 0[/tex], you have a=1, b=(-1), c=(-2)
 
  • #7
Yes ofc, but from there, where next? This is new to me I have never used inequalities in a graph let alone finding and solving question. I hardly even understand the question, it says soultions... suggesting more than one answer.
 
  • #8
ofc? Are you referring to me?
 
  • #9
zgozvrm said:
ofc? Are you referring to me?
I think that might be text-speak for "of course."
 
  • #10
greener1993 said:
Yes ofc, but from there, where next? This is new to me I have never used inequalities in a graph let alone finding and solving question. I hardly even understand the question, it says soultions... suggesting more than one answer.

For f(x)<0, solve for f(x)=0

The quadratic equation may or may not give you 2 answers.

In the case that it does give you 2 values for x, let's say your answers are x=a and x=b
That would mean that f(x)=(x-a)(x-b)

And, since we're trying to solve for f(x)<0, we have (x-a)(x-b)<0
Obviously, values of x=a and x=b will result in f(x)=0

What values of x will give you negative results?
In other words, for the inequality (x-a)(x-b)<0, what values make one factor negative and the other factor positive?
 
  • #11
not a clue. i worked out X is 2 or x=-1. I don't see how this relates to my question when there in the complex form of x(x-2)<2-x.

I had surgery on the 12th and i missed a lot of this. I am sorry but I seriously don't know what's happening.
 
  • #12
Just had a think about it and look some stuff online. For the first one do you factorise x^2-x-2 = (x-2)(x+1) so x=2 or x=-1. I then drew this on a graph to find out when y is greater than 0. Which would be -1>x>2. I don't know if this is right as i haven't used soultion a in the question :S
 
  • #13
When hallsofivy said it would be easier if you noticed that 2-x=-(x-2), this would be helpful because we have [tex]x(x-2)=-(x-2)[/tex] which means you can factorize it straight away. Say if we let x-2=A then we have [tex]xA=-A[/tex] then we add A to both sides and factorize giving [tex]A(x+1)=0[/tex] and then substituting back, [tex](x-2)(x+1)=0[/tex]. Of course there is no need to substitute for A, you can do it all without subbing if you understand the process.

So we have the answer to 1A)
[tex]x(x-2)=2-x[/tex]
[tex](x-2)(x+1)=0[/tex]

This is a parabola with roots -1 and 2 which is concave upwards (in other words, it grows very large for large positive and negative values of x, since the coefficient of the x2 term is positive).

B) I) for what values of x is [tex]x(x-2)<2-x[/tex] which - using the result from part A - means the same thing as for what values of x is [tex](x-2)(x+1)<0[/tex]. Now all you have to do is look at your parabola and choose the range of values for which the graph is under the x-axis (which is the same as being less than zero). You can always check to see if you're right by substituting any appropriate value of x back into the original question [tex]x(x-2)<2-x[/tex].

II) and III) are simple, you should be able to do those.

IV) In the same way we were asked to solve for B) I) we needed to move everything to one side and factorize first (or use the quadratic formula) to know the roots of the equation before proceeding. Then do the same that I've shown in the previous question.
 
  • #14
greener1993 said:
I then drew this on a graph to find out when y is greater than 0. Which would be -1>x>2.

Probably just a typo, but you can't have (-1) > X > 2

That means that x is both less than -1 AND greater than 2 - not possible.
Surely you meant (-1) < X < 2


Typo or not, mathematics is all about the details...
 
  • #15
zgozvrm said:
Probably just a typo, but you can't have (-1) > X > 2

That means that x is both less than -1 AND greater than 2 - not possible.
Surely you meant (-1) < X < 2


Typo or not, mathematics is all about the details...

Not just that, but also a minor detail that's been missed is that [tex](x-2)(x+1)>0[/tex] for [tex]x<-1, x>2[/tex]

Oh the details... :tongue:
 
  • #16
zgozvrm said:
For f(x)<0, solve for f(x)=0

The quadratic equation may or may not give you 2 answers.

In the case that it does give you 2 values for x, let's say your answers are x=a and x=b
That would mean that f(x)=(x-a)(x-b)

And, since we're trying to solve for f(x)<0, we have (x-a)(x-b)<0
Obviously, values of x=a and x=b will result in f(x)=0

What values of x will give you negative results?
In other words, for the inequality (x-a)(x-b)<0, what values make one factor negative and the other factor positive?

greener1993 said:
not a clue. i worked out X is 2 or x=-1. I don't see how this relates to my question when there in the complex form of x(x-2)<2-x.

Just as you did with x(x-2) = 2-x, rearrange the inequality so that you have zero on one side:
[itex]x(x-2) < 2-x[/tex]
[itex]x^2 -2x < 2-x[/tex]
[itex]x^2 - 2x -2 + x < 0[/tex]
[itex]x^2 - x - 2 < 0[/tex]

You already solved for x for the case where [itex]x^2 - x - 2 = 0[/tex]
You have x = 2 and x = -1



So for (x - 2)(x + 1) < 0, consider the following 3 cases:
1) values of x < -1
2) values of x between -1 and 2 (-1 < x < 2)
3) values of x > 2

For which case or cases is the inequality true?
There's your answer!


To start: for case #3, it should be easy to see that for any x > 2, you will have a positive result and therefore the inequality fails.
 
  • #17
zgozvrm said:
Just as you did with x(x-2) = 2-x, rearrange the inequality so that you have zero on one side:
[itex]x(x-2) < 2-x[/tex]
[itex]x^2 -2x < 2-x[/tex]
[itex]x^2 - 2x -2 + x < 0[/tex]
[itex]x^2 - x - 2 < 0[/tex]
As I said before, I don't think that's a good idea. With x(x- 2)= 2- x= -(x- 2), either x= 2 (so that both sides are 0) or it is not and we can divide by x- 2: x= -1. That gives the solutions immediately.

You already solved for x for the case where [itex]x^2 - x - 2 = 0[/tex]
You have x = 2 and x = -1



So for (x - 2)(x + 1) < 0, consider the following 3 cases:
1) values of x < -1
2) values of x between -1 and 2 (-1 < x < 2)
3) values of x > 2

For which case or cases is the inequality true?
There's your answer!


To start: for case #3, it should be easy to see that for any x > 2, you will have a positive result and therefore the inequality fails.
Or- the values x= 2 and x= -1, for which x(x-2) = 2-x, separate x(x-2)< 2- x from x(x-2)> 2- x. So we need only try one value of x in each interval:
1) x< -1. Take x= -2. (-2)(-2-2)= (-2)(-4)= 8> 2-(-2)= 4 so the inequality is false for all x less than -1.
2) -1< x< 2. Take x= 0. (0)(-2-0)= 0< 2- 0= 2 so the inequality is true for all x between -1 and 2.
3) x> 2. Take x= 3. (3)(3- 2)= 3(1)= 3< 2- 3= -1 so the inequality is false for all x greater than 2.
 
  • #18
HallsofIvy said:
As I said before, I don't think that's a good idea. With x(x- 2)= 2- x= -(x- 2), either x= 2 (so that both sides are 0) or it is not and we can divide by x- 2: x= -1. That gives the solutions immediately.

Why not? He's already solved for X in part I: X1 = -1 and X2 = 2
That means the factors of [itex]x^2 - x - 2[/tex] are (X - X1) and (X - X2)
So you instantly have (X + 1)(X - 2) < 0



HallsofIvy said:
Or- the values x= 2 and x= -1, for which x(x-2) = 2-x, separate x(x-2)< 2- x from x(x-2)> 2- x. So we need only try one value of x in each interval:
1) x< -1. Take x= -2. (-2)(-2-2)= (-2)(-4)= 8> 2-(-2)= 4 so the inequality is false for all x less than -1.
2) -1< x< 2. Take x= 0. (0)(-2-0)= 0< 2- 0= 2 so the inequality is true for all x between -1 and 2.
3) x> 2. Take x= 3. (3)(3- 2)= 3(1)= 3< 2- 3= -1 so the inequality is false for all x greater than 2.

... then given that inequality, plugging in the numbers for each case is much easier.
In fact, one should be able to solve it by inspection at this point:

The factors (X - 2) and (X + 1) must have opposite signs in order for the product to be negative. Obviously, any value for X > 2 will result in a positive times a positive. Any value for X < -1 will result in a negative times a negative. Any value for X between those values will result in a negative times a positive, which is what we're looking for.
 
  • #19
When I was learning about quadratics inequalities, it was much easier to draw the graph of y=f(x) when trying to figure out when f(x)<0. Maybe the OP would prefer that way too.
 
  • #20
Mentallic said:
When I was learning about quadratics inequalities, it was much easier to draw the graph of y=f(x) when trying to figure out when f(x)<0. Maybe the OP would prefer that way too.

Yeah, but try that with a problem like [itex]4x^2 - 9x - 27 < 2[/tex]
 
  • #21
zgozvrm said:
Yeah, but try that with a problem like [itex]4x^2 - 9x - 27 < 2[/tex]

Rearranged to [tex]4x^2-9x-29<0[/tex]
Finding the roots with the quadratic formula, let them be x1 and x2
The parabola is concave up, so then [tex]x_1<x<x_2[/tex]

I don't see the point you were trying to make here. You need to find the roots either way and I just suggested that the student might find the picture of the parabola more simple to understand than taking cases - which is more work anyway.
 
  • #22
Mentallic said:
I don't see the point you were trying to make here. You need to find the roots either way and I just suggested that the student might find the picture of the parabola more simple to understand than taking cases - which is more work anyway.

The point I'm making is that the answer to the problem I posed is approximately -1.793 < x < 4.043, or more accurately:

[tex]\frac{9 - \sqrt{545}}{8} < x < \frac{9 + \sqrt{545}}{8}[/tex]

How can you read that from a graph?
 
  • #23
... besides, I can enter the function in either Excel or my $20 scientific calculator and try values below x1, above x2 and between x1 & x2 faster than I can draw a graph.
 
  • #24
zgozvrm said:
The point I'm making is that the answer to the problem I posed is approximately -1.793 < x < 4.043, or more accurately:

[tex]\frac{9 - \sqrt{545}}{8} < x < \frac{9 + \sqrt{545}}{8}[/tex]

How can you read that from a graph?

zgozvrm said:
... besides, I can enter the function in either Excel or my $20 scientific calculator and try values below x1, above x2 and between x1 & x2 faster than I can draw a graph.

:uhh:

You find the roots, then you draw an approximate graph, which should then show you where the function is less than zero or more than zero. I never mentioned anything about approximating the roots. I was only providing a (in my opinion) more simple approach to figuring out the range of x values for which the quadratic is less than zero.
 

1. What are quadratic inequalities?

Quadratic inequalities are mathematical equations that involve a quadratic expression, where the highest power of the variable is 2, and an inequality symbol, such as <, >, ≤, or ≥. These equations can be solved to find the values of the variable that satisfy the inequality.

2. How do you graph quadratic inequalities?

To graph a quadratic inequality, first graph the corresponding quadratic equation. Then, depending on the inequality symbol, shade the region above or below the graph to represent the solutions. If the inequality includes an equal sign, the boundary line should be included in the shaded region.

3. What is the difference between a quadratic equation and a quadratic inequality?

A quadratic equation is an equation that is set equal to 0 and can be solved to find specific values of the variable. A quadratic inequality, on the other hand, involves an inequality symbol and can have multiple solutions that satisfy the inequality.

4. How do you solve quadratic inequalities algebraically?

To solve quadratic inequalities algebraically, first bring all terms to one side of the inequality, leaving 0 on the other side. Factor the quadratic expression and set each factor equal to 0. Solve for the variable in each equation and then use a number line or interval notation to represent the solutions.

5. Can you solve quadratic inequalities using the quadratic formula?

Yes, the quadratic formula can also be used to solve quadratic inequalities. After setting the inequality equal to 0, plug in the coefficients of the quadratic expression into the quadratic formula. The solutions will be the x-intercepts of the corresponding quadratic equation, which can be used to graph the inequality.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
4
Views
606
  • Precalculus Mathematics Homework Help
Replies
5
Views
634
  • Precalculus Mathematics Homework Help
Replies
5
Views
721
  • Precalculus Mathematics Homework Help
Replies
9
Views
706
  • Precalculus Mathematics Homework Help
Replies
7
Views
700
  • Precalculus Mathematics Homework Help
Replies
13
Views
164
  • Precalculus Mathematics Homework Help
Replies
11
Views
326
  • Precalculus Mathematics Homework Help
Replies
1
Views
459
  • Precalculus Mathematics Homework Help
Replies
6
Views
528
  • Precalculus Mathematics Homework Help
Replies
5
Views
762
Back
Top