# B Solution set of a square root

1. May 8, 2017

### mech-eng

When we find solution set of an equation inside a square root why we should assume that inside of square root should be equal to or greater than zero? For example $\sqrt{5x-4}$.

How can I use here equal to or greater than zero symbol?

Thank you.

2. May 8, 2017

### BvU

As long as you are restricting yourself to real numbers, the expression under the square root sign has to be non-negative, so you can square if you want, but then you need to add & $5x -4 \ge 0$

3. May 8, 2017

### mech-eng

For only real number, I cannot see that reason which restricting inside a square root being $\ge 0$? I think it is not easy to see this. Yes if I take the square of both sides it is $5x-4\ge 0$. This is an easy step but can it give any explanation?

Thank you.

4. May 8, 2017

### BvU

There is no real number $x$ for which $x^2 < 0$

5. May 8, 2017

### mech-eng

But there can be $real\ numbers$ which can make $5x-4$ negative so when putting $5x-4$ inside a square root why should these values be neglected? I cannot understand this part easily.

Thank you.

6. May 8, 2017

### Staff: Mentor

Because sometimes a complex solution is not acceptable. If you have $t = \sqrt{5x-4}$, where $t$ is time, then only real solutions are physically acceptable.

7. May 8, 2017

### mech-eng

Without thinking both complex numbers and meaningful physical situation, i.e from respect of pure real numbers, what is the reason?

Thank you.

8. May 8, 2017

### Staff: Mentor

If you restrict the codomain of a function to the real numbers, then that may restrict its domain.

9. May 8, 2017

### Staff: Mentor

Because for $\sqrt{5x - 4}$ to be real, $5x - 4$ must be greater than or equal to 0. Otherwise (if $5x - 4 < 0$), you'll be taking the square root of a negative number.

10. May 8, 2017

### mech-eng

Yes this explains it but requires more thinking. Would you please check if this explanation expanding it correct?
"If $5x-4$ negative real number and if we put it into square root $\sqrt{5x-4}$ cannot a real number because when we take square of $\sqrt{5x-4}^2$ this will be $5x-4$, a negative real number, which is imposibble.

Thank you.

Last edited: May 8, 2017
11. May 8, 2017

### FactChecker

You should not assume that 5x-4 is ≥0. You should state that your answers are only valid for 5x-4 ≥0. That is significantly different.
You can allow square roots of negative numbers if you are willing to talk about imaginary and complex numbers.

PS. When you are editing a post, the ≥ symbol (and many others) is available by clicking the Σ symbol at the top right.

12. May 8, 2017

### mech-eng

But I regard only reel numbers, do not assume complex numbers.

Thank you.

13. May 8, 2017

### FactChecker

I understand. Just a little "word-smithing". I don't like using the word "assume" when that may not be true. It's better to state a restriction that is required to make something valid than to assume it.
Like: $\sqrt{5x-4} = etc... ; x ≥ 4/5$
Then you should also keep track of those restrictions when you give your final answer.

14. May 8, 2017

### mech-eng

But does not #10 explains corrrectly why this restriction should be?

Thank you.

15. May 8, 2017

### FactChecker

Yes, it does. As long as you keep in mind that it is a restriction that you should keep track of, rather than an assumption that you don't need to worry about.

16. May 8, 2017

### FactChecker

When you are editing a post, the ≥ symbol (and many others) is available by clicking the Σ symbol at the top right.

17. May 8, 2017

### Staff: Mentor

You're making it much more complicated that it needs to be.
From post #1,
The usual square root function, the one that produces a real number, is defined only for expressions whose values are ≥ 0. This means that $5x - 4 \ge 0$ must be true, or equivalently, $x \ge \frac 4 5$. In this case, if 5x - 4 < 0, its square root will not be a real number.​
That's all you really need to say.