Finding the Range of x Values for Quadratic Inequalities

AI Thread Summary
The discussion centers on solving quadratic inequalities and simultaneous equations. Participants clarify the process of solving the equation x(x-2) = 2-x, emphasizing the importance of factoring and identifying intervals where the inequality holds true. There is confusion regarding the simultaneous equations, with a focus on ensuring correct substitutions and signs. Simplification of expressions, particularly involving square roots, is also addressed, with guidance on using the quadratic formula. Overall, the conversation highlights the need for careful attention to detail in mathematical problem-solving.
greener1993
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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
 

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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.
 
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
\frac{4 \pm \sqrt{20}}{2}
or
4 \pm \frac{\sqrt{20}}{2}
or
4 \pm \sqrt{\frac{20}{2}}
 
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 x^2- x- 2= 0 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 x\ne 2, 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
\frac{4\pm\sqrt{20}}{2}= \frac{4\pm\sqrt{4(5)}}{2}= \frac{4\pm 2\sqrt{5}}{2}= 2\pm\sqrt{5}
You cancel the "2" in front of the square root.
\frac{\sqrt{5}}{2}\ne \sqrt{\frac{5}{2}}

Thank you for your help
 
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 :(
 
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 ax^2 + bx + c = 0[/tex], solve for x by;<br /> <br /> x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}<br /> <br /> <br /> So, in the case of x^2 - x -2 = 0[/tex], you have a=1, b=(-1), c=(-2)
 
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.
 
ofc? Are you referring to me?
 
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 x(x-2)=-(x-2) which means you can factorize it straight away. Say if we let x-2=A then we have xA=-A then we add A to both sides and factorize giving A(x+1)=0 and then substituting back, (x-2)(x+1)=0. 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)
x(x-2)=2-x
(x-2)(x+1)=0

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 x(x-2)&lt;2-x which - using the result from part A - means the same thing as for what values of x is (x-2)(x+1)&lt;0. 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 x(x-2)&lt;2-x.

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 (x-2)(x+1)&gt;0 for x&lt;-1, x&gt;2

Oh the details... :-p
 
  • #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:
x(x-2) &lt; 2-x[/tex]<br /> x^2 -2x &amp;lt; 2-x[/tex]&lt;br /&gt; x^2 - 2x -2 + x &amp;amp;lt; 0[/tex]&amp;lt;br /&amp;gt; x^2 - x - 2 &amp;amp;amp;lt; 0[/tex]&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; You already solved for x for the case where x^2 - x - 2 = 0[/tex]&amp;amp;amp;lt;br /&amp;amp;amp;gt; You have x = 2 and x = -1&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; So for (x - 2)(x + 1) &amp;amp;amp;amp;lt; 0, consider the following 3 cases:&amp;amp;amp;lt;br /&amp;amp;amp;gt; 1) values of x &amp;amp;amp;amp;lt; -1&amp;amp;amp;lt;br /&amp;amp;amp;gt; 2) values of x between -1 and 2 (-1 &amp;amp;amp;amp;lt; x &amp;amp;amp;amp;lt; 2)&amp;amp;amp;lt;br /&amp;amp;amp;gt; 3) values of x &amp;amp;amp;amp;gt; 2&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; For which case or cases is the inequality true?&amp;amp;amp;lt;br /&amp;amp;amp;gt; There&amp;amp;amp;amp;#039;s your answer!&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; To start: for case #3, it should be easy to see that for any x &amp;amp;amp;amp;gt; 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:
x(x-2) &lt; 2-x[/tex]<br /> x^2 -2x &amp;lt; 2-x[/tex]&lt;br /&gt; x^2 - 2x -2 + x &amp;amp;lt; 0[/tex]&amp;lt;br /&amp;gt; x^2 - x - 2 &amp;amp;amp;lt; 0[/tex]
&amp;amp;lt;br /&amp;amp;gt; As I said before, I don&amp;amp;amp;#039;t think that&amp;amp;amp;#039;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.&amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;br /&amp;amp;gt; &amp;amp;lt;blockquote data-attributes=&amp;amp;quot;&amp;amp;quot; data-quote=&amp;amp;quot;&amp;amp;quot; data-source=&amp;amp;quot;&amp;amp;quot; class=&amp;amp;quot;bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch&amp;amp;quot;&amp;amp;gt; &amp;amp;lt;div class=&amp;amp;quot;bbCodeBlock-content&amp;amp;quot;&amp;amp;gt; &amp;amp;lt;div class=&amp;amp;quot;bbCodeBlock-expandContent js-expandContent &amp;amp;quot;&amp;amp;gt; You already solved for x for the case where x^2 - x - 2 = 0[/tex]&amp;amp;amp;lt;br /&amp;amp;amp;gt; You have x = 2 and x = -1&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; So for (x - 2)(x + 1) &amp;amp;amp;amp;lt; 0, consider the following 3 cases:&amp;amp;amp;lt;br /&amp;amp;amp;gt; 1) values of x &amp;amp;amp;amp;lt; -1&amp;amp;amp;lt;br /&amp;amp;amp;gt; 2) values of x between -1 and 2 (-1 &amp;amp;amp;amp;lt; x &amp;amp;amp;amp;lt; 2)&amp;amp;amp;lt;br /&amp;amp;amp;gt; 3) values of x &amp;amp;amp;amp;gt; 2&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; For which case or cases is the inequality true?&amp;amp;amp;lt;br /&amp;amp;amp;gt; There&amp;amp;amp;amp;#039;s your answer!&amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; &amp;amp;amp;lt;br /&amp;amp;amp;gt; To start: for case #3, it should be easy to see that for any x &amp;amp;amp;amp;gt; 2, you will have a positive result and therefore the inequality fails. &amp;amp;lt;/div&amp;amp;gt; &amp;amp;lt;/div&amp;amp;gt; &amp;amp;lt;/blockquote&amp;amp;gt;&amp;amp;amp;lt;br /&amp;amp;amp;gt; Or- the values x= 2 and x= -1, for which x(x-2) = 2-x, &amp;amp;amp;lt;b&amp;amp;amp;gt;separate&amp;amp;amp;lt;/b&amp;amp;amp;gt; x(x-2)&amp;amp;amp;amp;lt; 2- x from x(x-2)&amp;amp;amp;amp;gt; 2- x. So we need only try one value of x in each interval:&amp;amp;amp;lt;br /&amp;amp;amp;gt; 1) x&amp;amp;amp;amp;lt; -1. Take x= -2. (-2)(-2-2)= (-2)(-4)= 8&amp;amp;amp;amp;gt; 2-(-2)= 4 so the inequality is false for &amp;amp;amp;lt;b&amp;amp;amp;gt;all&amp;amp;amp;lt;/b&amp;amp;amp;gt; x less than -1.&amp;amp;amp;lt;br /&amp;amp;amp;gt; 2) -1&amp;amp;amp;amp;lt; x&amp;amp;amp;amp;lt; 2. Take x= 0. (0)(-2-0)= 0&amp;amp;amp;amp;lt; 2- 0= 2 so the inequality is true for &amp;amp;amp;lt;b&amp;amp;amp;gt;all&amp;amp;amp;lt;/b&amp;amp;amp;gt; x between -1 and 2. &amp;amp;amp;lt;br /&amp;amp;amp;gt; 3) x&amp;amp;amp;amp;gt; 2. Take x= 3. (3)(3- 2)= 3(1)= 3&amp;amp;amp;amp;lt; 2- 3= -1 so the inequality is false for &amp;amp;amp;lt;b&amp;amp;amp;gt;all&amp;amp;amp;lt;/b&amp;amp;amp;gt; 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 x^2 - x - 2[/tex] are (X - X1) and (X - X2)<br /> So you instantly have (X + 1)(X - 2) &lt; 0<br /> <br /> <br /> <br /> <blockquote data-attributes="" data-quote="HallsofIvy" data-source="post: 2955618" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> HallsofIvy said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Or- the values x= 2 and x= -1, for which x(x-2) = 2-x, <b>separate</b> x(x-2)&lt; 2- x from x(x-2)&gt; 2- x. So we need only try one value of x in each interval:<br /> 1) x&lt; -1. Take x= -2. (-2)(-2-2)= (-2)(-4)= 8&gt; 2-(-2)= 4 so the inequality is false for <b>all</b> x less than -1.<br /> 2) -1&lt; x&lt; 2. Take x= 0. (0)(-2-0)= 0&lt; 2- 0= 2 so the inequality is true for <b>all</b> x between -1 and 2. <br /> 3) x&gt; 2. Take x= 3. (3)(3- 2)= 3(1)= 3&lt; 2- 3= -1 so the inequality is false for <b>all</b> x greater than 2. </div> </div> </blockquote><br /> ... then given that inequality, plugging in the numbers for each case is <b>much</b> easier.<br /> In fact, one should be able to solve it by inspection at this point:<br /> <br /> The factors (X - 2) and (X + 1) must have opposite signs in order for the product to be negative. Obviously, any value for X &gt; 2 will result in a positive times a positive. Any value for X &lt; -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&#039;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 4x^2 - 9x - 27 &lt; 2[/tex]
 
  • #21
zgozvrm said:
Yeah, but try that with a problem like 4x^2 - 9x - 27 &lt; 2[/tex]
<br /> <br /> Rearranged to 4x^2-9x-29&amp;lt;0<br /> Finding the roots with the quadratic formula, let them be x<sub>1</sub> and x<sub>2</sub><br /> The parabola is concave up, so then x_1&amp;lt;x&amp;lt;x_2<br /> <br /> I don&#039;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:

\frac{9 - \sqrt{545}}{8} &lt; x &lt; \frac{9 + \sqrt{545}}{8}

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:

\frac{9 - \sqrt{545}}{8} &lt; x &lt; \frac{9 + \sqrt{545}}{8}

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.

:rolleyes:

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.
 

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