Solving Inequality: sqrt(x-1)+sqrt(x-3)+8>0

[yoleven]

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.

 

[futurebird]

This is your problem right?

$$\sqrt{x-1}+\sqrt{x-3}+8 > 0$$

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

$$\sqrt{x-1}+\sqrt{x-3}+8$$

Only, think of $$\sqrt{x-1}=a$$ $$\sqrt{x-3}=b$$ 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:

$$\sqrt{x-1}+\sqrt{x-3}+8 > 0$$

we know that

$$\sqrt{x-1}+\sqrt{x-3}> -8$$

Well, when are square roots ever negative?
So, what restrictions should you place on X?

 

[yoleven]

This is your problem right?

$$\sqrt{x-1}+\sqrt{x-3}+8 > 0$$

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

$$\sqrt{x-1}+\sqrt{x-3}+8$$

Only, think of $$\sqrt{x-1}=a$$ $$\sqrt{x-3}=b$$ 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:

$$\sqrt{x-1}+\sqrt{x-3}+8 > 0$$

we know that

$$\sqrt{x-1}+\sqrt{x-3}> -8$$

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?

 

[dynamicsolo]

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

$$\sqrt{x-1}+\sqrt{x-3}+8 > 0$$

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 $$x - 1 \geq 0$$ and the second term, for $$x - 3 \geq 0$$. The intersection of those sets is $$x \geq 3$$. You can make a similar argument for the inequality you suggested, for which you found $$x \geq 9$$.

 

[spideyunlimit]

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)

 

[rocomath]

Just find the domain and you've got your answer. You don't need to square or move anything around.

 

[yoleven]

Thank you.