Why can rhs be lower than 0 if there is a square root on lhs?

[Loki123]

Why are there two cases in the following problem if the RHS can only be positive since LHS is square root and square root can only be positive?

 

[lev888]

Why are there two cases in the following problem if the RHS can only be positive since LHS is square root and square root can only be positive? View attachment 29149

Let's call the square root R. We have
R > -x, -x>=0
R > -x, -x<0

Yes, R is positive (rather non-negative). But... these are still different cases:

R > negative value => true for ANY x, such that -x < 0 and the expression under root is not negative.

R > non-negative value => need to solve.

 

[Loki123]

Let's call the square root R. We have
R > -x, -x>=0
R > -x, -x<0

Yes, R is positive (rather non-negative). But... these are still different cases:

R > negative value => true for ANY x, such that -x < 0 and the expression under root is not negative.

R > non-negative value => need to solve.

I am sorry, could you perhaps try explaining that again, I am having a hard time understanding.

 

[lev888]

I am sorry, could you perhaps try explaining that again, I am having a hard time understanding.

You wrote:
"Why are there two cases in the following problem if the RHS can only be positive since LHS is square root and square root can only be positive?"

Why do you think that RHS can only be positive? We only know that LHS is greater than RHS. If we knew that RHS was positive, then we could conclude that LHS had to be positive.

 

[Dr.Peterson]

Why are there two cases in the following problem if the RHS can only be positive since LHS is square root and square root can only be positive? View attachment 29149

But the RHS can be positive. For example, if x=1, the LHS is 1 and the RHS is -1; the inequality is true.

Perhaps you are confusing this with an equation.

 

[pka]

Why are there two cases in the following problem if the RHS can only be positive since LHS is square root and square root can only be positive? View attachment 29149

It would been nice if you had given us some idea as to the level of mathematics you are studying. It seems that you may not be as comfortable with inequalities as this problem requires. If solving inequalities we often suggest that tye expression related to zero.
Thus $\sqrt{x^2-x+1}>-x$ first becomes $\sqrt{x^2-x+1}+x>0$.
The we would encourage the use of a graphing utility to examine the graphs.
Look at this. We can that for all $x,~x^2-x+1>0$ thus for every $x,~\sqrt{x^2-x+1}>0$
Look at the graph of $\sqrt{x^2-x+1}>-x$ HERE.
What do you see? Can you explain the solution?
Please post what you find.
$$$$$$$$

 

[Loki123]

You wrote:
"Why are there two cases in the following problem if the RHS can only be positive since LHS is square root and square root can only be positive?"

Why do you think that RHS can only be positive? We only know that LHS is greater than RHS. If we knew that RHS was positive, then we could conclude that LHS had to be positive.

Because LHS is square root and square root cannot be negative. So RHS should be =>0

 

[Loki123]

It would been nice if you had given us some idea as to the level of mathematics you are studying. It seems that you may not be as comfortable with inequalities as this prom requires. If solving inequalities we often suggest that tye expression related to zero.
Thus $\sqrt{x^2-x+1}>-x$ first becomes $\sqrt{x^2-x+1}+x>0$.
The we would encourage the use of a graphing utility to examine the graphs.
Look at this. We can that for all $x,~x^2-x+1>0$ thus for every $x,~\sqrt{x^2-x+1}>0$
Look at the graph of $\sqrt{x^2-x+1}>-x$ HERE.
What do you see? Can you explain the solution?
Please post what you find.
$$$$$$$$

The x was on the lhs, i switched it over to the rhs to solve the inequality.

 

[lev888]

Because LHS is square root and square root cannot be negative. So RHS should be =>0

Please explain "so". What is your reasoning?

 

[Dr.Peterson]

But the RHS can be negative. For example, if x=1, the LHS is 1 and the RHS is -1; the inequality is true.

Perhaps you are confusing this with an equation.

This is a correction to what I wrote before. Do you see that this contradicts your claim? That should make you reevaluate your reasoning.

 

[pka]

The x was on the lhs, i switched it over to the rhs to solve the inequality.

Well that means that you are in a worst place than I imagined you to be.
If the question was given as $\sqrt{x^2-x+1}+x>0$ then the whole point is to see that the inequality is true for all $x$.
Did you study the graphs in the links I posted? This is really a visual exercise. see here

$$$$$$

 

[Steven G]

The left hand side equals whatever the right side equal and the rhs equals whatever the lhs equals. That is a property of the equal sign.

Now, the left hand side can't be negative--it can only be 0 or positive. The same must be true for the rhs- it can only be 0 or positive.

Hence, why can the rhs be lower than 0? The rhs CAN NOT be lower than 0? Why do you think otherwise?

​

 

[Dr.Peterson]

The left hand side equals whatever the right side equal and the rhs equals whatever the lhs equals. That is a property of the equal sign.

Now, the left hand side can't be negative--it can only be 0 or positive. The same must be true for the rhs- it can only be 0 or positive.

Hence, why can the rhs be lower than 0? The rhs CAN NOT be lower than 0? Why do you think otherwise?

​

You are aware that it is an inequality, not an equation, right?

All we know (initially) is that the RHS is less than some positive number. That doesn't rule out any negative numbers.

Did you study the graphs in the links I posted? This is really a visual exercise.

True, it can be seen graphically; but the question is about a step in an algebraic solution of the inequality. Presumably the next step is to square both sides (in the first case), and to observe that the second case is always true, because any positive number is greater than any negative number. The OP seems to think that the second case is never true.

 

[Steven G]

You are aware that it is an inequality, not an equation, right?

All we know (initially) is that the RHS is less than some positive number. That doesn't rule out any negative numbers.


True, it can be seen graphically; but the question is about a step in an algebraic solution of the inequality. Presumably the next step is to square both sides (in the first case), and to observe that the second case is always true, because any positive number is greater than any negative number. The OP seems to think that the second case is never true.

I based my post on the title of the post.

 

[lookagain]

I based my post on the title of the post.

And there are false, misleading, ambiguous headlines of articles that
contrast/contradict with the content of the body of articles, or the
clickbait titles of videos that do not deliver what they promise.

 

[Dr.Peterson]

I based my post on the title of the post.

How does the title, "Why can rhs be lower than 0 if there is a square root on lhs?" imply the two sides are equal?

As a mentor of mine in software engineering long ago used to say, "Never. Assume. Anything." In solving a math problem, it's always essential to read the problem, think about it, and then read again. The same is true in answering a mere question.

(This does not imply that I never misread anything!)

 

[Loki123]

But the RHS can be positive. For example, if x=1, the LHS is 1 and the RHS is -1; the inequality is true.

Perhaps you are confusing this with an equation.

Would you mind checking this post out? I tried to word it a bit better.

Equations and inequations with square root

I have posted a few problems that gave me the same issue. Unfortunately, I haven't been able to understand the reasoning on why they are done the way they are. So I decided to sum up all my problems regarding that issue here in hopes to understand it. The following picture includes two equations...

www.freemathhelp.com