Why do these two inequalities give these results?

[itsrayex]

I have two questions:

1. Photomath tells me the first inequality is impossible because the left side is always bigger than zero. First of all, I thought the way to solve inequalities of second degree was to use the specific formula, not by just looking at it. If you actually use the formula on inequality nr20 it yelds a numeric result, so I don't get it: when do we use the formula?

2. Assuming nr20 was impossible, then shouldn't the other exercise also yeld the same result? The left side will never be less, nor equal to zero, because since it has a positive exponent whatever sign the x has it's always going to be turned into a + sign and therefore get added to the 3, that also has a + sign. Since both are plus and 3 is not zero, how come is the result x=-3 and not impossible?

Thank you for your attention

 

[blamocur]

the left side is always bigger than zero.

Not necessary always bigger, but never less than zero.

I thought the way to solve inequalities of second degree was to use the specific formula

You might use a specific formula in general case, but in this particular case you can figure out the answer without formulas. BTW, which formula yielded a different result?

nor equal to zero,

Why not ? What about [imath]x=-3[/imath]?

 

[pka]

Every non-zero real number squared is positive( not negative). That is why #20 is impossible.
#21` is possible only for zero. (x=-3)

 

[JeffM]

I am not sure what method you are talking about. No method overturns fundamentals. If you square any real number, you get a non-negative number. And complex numbers do not directly have an order relationship. You can impose such an order relationship on the complex numbers through norms, but in that case, the norm of any complex number > the norm of 0.

 

[BigBeachBanana]

I am not sure what method you are talking about. No method overturns fundamentals. If you square any real number, you get a non-negative number. And complex numbers do not directly have an order relationship. You can impose such an order relationship on the complex numbers through norms, but in that case, the norm of any complex number > the norm of 0.

The question was posted under beginning algebra, complex analysis might not be appropriate

 

[Dr.Peterson]

View attachment 30762
I have two questions:

1. Photomath tells me the first inequality is impossible because the left side is always bigger than zero. First of all, I thought the way to solve inequalities of second degree was to use the specific formula, not by just looking at it. If you actually use the formula on inequality nr20 it yelds a numeric result, so I don't get it: when do we use the formula?

Please show your actual work, so we can see what "formula" you used, and where you made a mistake with it. The methods I know of will yield the correct answer, but that may involve rejecting extraneous solutions.

 

[Steven G]

(4x-5)^2 < 0

No matter what number x is, when you multiply it by 4 you get some (other) number. Then you subtract 5 and get another number. No matter what number you have now when you square it you'll have 0 or greater than 0. Never less than 0.

(x+3)^2 < 0 means (x+3)^2 < 0 or (x+3)^2 = 0
Like nr20, (x+3)^2 < 0 has no solution.
However, (x+3)^2 = 0 has x=-3 as a solution.

How would you handle (x+3)^2 > 0?