I don't use the equation solver right or its a bug?

[hadasz]

Hi,

I have this inequation:

\(\displaystyle \dfrac{2x^2\, +\, 9x\, +\, 4}{(x\, -\, 3)^2}\, \leq\, 0\)

The answer is: -4 < x < -0.5, right?

But the online equation solver (using the "Algebra" tab) gives: x<=0

Did I do something wrong?

Thanks for your help.

 

[stapel]

Reply to post.

\(\displaystyle \dfrac{2x^2\, +\, 9x\, +\, 4}{(x\, -\, 3)^2}\, \leq\, 0\)

The answer is: -4 < x < -0.5, right?

That's what I'm getting, yes.

the online equation solver (using the "Algebra" tab) gives: x<=0

Did I do something wrong?

Yes. You chose "convert to interval notation" rather than "solve the inequality". As such, the solver did exactly what you had asked it to do. It took the inequality "(this) is less than or equal to (zero)" and converted it to "(this) is true on the interval (-infinity through zero)". But this says nothing about the values of x which make the inequality true. :wink:

 

[Ishuda]

Hi,

I have this inequation:

\(\displaystyle \dfrac{2x^2\, +\, 9x\, +\, 4}{(x\, -\, 3)^2}\, \geq\, 0\)

The answer is: -4 < x < -0.5, right?

But the online equation solver (using the "Algebra" tab) gives: x<=0

Did I do something wrong?

Thanks for your help.

Excuse me, but do you mean the complement of that interval?
\(\displaystyle \dfrac{2*10^2\, +\, 9*10\, +\, 4}{(10\, -\, 3)^2}\, =\, \frac{100\, +\, 90\, +4}{49}\, =\, \frac{194}{49}\, \ge\, 0\)

 

[hadasz]

solve the ineqality

That's what I'm getting, yes.


Yes. You chose "convert to interval notation" rather than "solve the inequality". As such, the solver did exactly what you had asked it to do. It took the inequality "(this) is less than or equal to (zero)" and converted it to "(this) is true on the interval (-infinity through zero)". But this says nothing about the values of x which make the inequality true. :wink:

Well... I don't have the option "solve the ineqality"
I tried some options: convert to interval notation, solve for x, ...

 

[hadasz]

yeh, your right...

Excuse me, but do you mean the complement of that interval?
\(\displaystyle \dfrac{2*10^2\, +\, 9*10\, +\, 4}{(10\, -\, 3)^2}\, =\, \frac{100\, +\, 90\, +4}{49}\, =\, \frac{194}{49}\, \ge\, 0\)

I made a typing mistake , in the original problem it was written <= and not >=

 

[stapel]

Well... I don't have the option "solve the ineqality"
I tried some options: ...solve for x, ...

And "solving for x" is the same as "solving the inequality for the solution values of x"; that is, "solving for x" IS "solving the inequality". I'm not sure how one could arrive that these sorts of homework problems and not be familiar with the basic terminology related to said problems. Is your instructor skipping things and not using a textbook of any sort? :shock: