• Support PF! Buy your school textbooks, materials and every day products Here!

Solving inequality

  • Thread starter yoleven
  • Start date
  • #1
78
0

Homework Statement



sqrt(x-1)+sqrt(x-3)+8>0

Homework Equations





The Attempt at a Solution


I thought I would square everything to get rid of the root signs...leaving me with
(x-1)+(x-3)+64>0
2x+60>0
x>30
ss:{xE[30,Infinity)}

However, the answer is supposed to be..
[3,infinity)
Please tell me what I am missing.
 

Answers and Replies

  • #2
270
0
This is your problem right?

[tex]\sqrt{x-1}+\sqrt{x-3}+8 > 0[/tex]

If you "square both sides" you will not get: (x-1)+(x-3)+64>0 !

Look at http://www.syvum.com/cgi/online/serve.cgi/squizzes/algebra/expand3.html" web page. The same rules shown there will apply if you square

[tex]\sqrt{x-1}+\sqrt{x-3}+8[/tex]

Only, think of [tex]\sqrt{x-1}=a[/tex] [tex]\sqrt{x-3}=b[/tex] and 8=c.

As you can see, when you square a trinomial you may end up with a big mess.

You can solve this problem by thinking about it for a bit. For this to be true:

[tex]\sqrt{x-1}+\sqrt{x-3}+8 > 0[/tex]

we know that

[tex]\sqrt{x-1}+\sqrt{x-3}> -8[/tex]

Well, when are square roots ever negative?
So, what restrictions should you place on X?
 
Last edited by a moderator:
  • #3
78
0
This is your problem right?

[tex]\sqrt{x-1}+\sqrt{x-3}+8 > 0[/tex]

If you "square both sides" you will not get: (x-1)+(x-3)+64>0 !

Look at http://www.syvum.com/cgi/online/serve.cgi/squizzes/algebra/expand3.html" web page. The same rules shown there will apply if you square

[tex]\sqrt{x-1}+\sqrt{x-3}+8[/tex]

Only, think of [tex]\sqrt{x-1}=a[/tex] [tex]\sqrt{x-3}=b[/tex] and 8=c.

As you can see, when you square a trinomial you may end up with a big mess.

You can solve this problem by thinking about it for a bit. For this to be true:

[tex]\sqrt{x-1}+\sqrt{x-3}+8 > 0[/tex]

we know that

[tex]\sqrt{x-1}+\sqrt{x-3}> -8[/tex]

Well, when are square roots ever negative?
So, what restrictions should you place on X?
Okay, I didn't recognize it as a trinomial. If x is less than 3 the equation won't work, right?
All I have to do is recognize what the minumum x must be so that I am not taking a sqrt of a negative number. Is that correct?

If the equation was sqrt(x-5)+sqrt(x-9)+14>0 would I be correct in stating that xE[9,infinity)
because then I am not taking the sqrt of a negative number. Is that all I have to do for this kind of problem?
 
Last edited by a moderator:
  • #4
dynamicsolo
Homework Helper
1,648
4
I'm surprised you hadn't gotten a reply by now...

Yes, you are correct. In this particular situation, since the square root operation only gives results that are positive or zero, the left-hand side of

[tex]
\sqrt{x-1}+\sqrt{x-3}+8 > 0
[/tex]

couldn't possibly be less than 8. So plainly the left-hand side is always greater than zero.

So what is the only problem you could run into? That the terms on the left-hand side be undefined. The first term is only defined for [tex]x - 1 \geq 0[/tex] and the second term, for [tex]x - 3 \geq 0[/tex]. The intersection of those sets is [tex]x \geq 3[/tex]. You can make a similar argument for the inequality you suggested, for which you found [tex]x \geq 9[/tex].
 
  • #5
sqrt(x-1) + sqrt(x-3) + 8 > 0

Well it will always be greater than 0, in fact always greater than or equal to 8 atleast, since the roots cannot be negative

now just consider the domain x >=3 ... because of the roots again
And that's your answer x E [3, infinity)
 
  • #6
1,752
1
Just find the domain and you've got your answer. You don't need to square or move anything around.
 
  • #7
78
0
Thank you.
 

Related Threads on Solving inequality

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
12
Views
2K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
14
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
1K
Top