Domain problem: how to find domain of y = sqrt[ (-3) / (x^2 - 5x + 4) ]

[AlexSendler100%]

How can I find the domain of this function?

$\qquad y = \sqrt{\dfrac{-3}{x^2-5x+4}\;}$

I mean, what in the square should be greater or equal to zero, but I can't divide the square because a square of -3 is impossible, and if I should make the inequality of all the expression to be great or equal to zero, how to solve this inequality

 

[stapel]

How can I find the domain of this function?

$\qquad y = \sqrt{\dfrac{-3}{x^2-5x+4}\;}$

I mean, what in the square should be greater or equal to zero...

For domains, start with "everything", and then remove any problems.

Polynomials, for instance, never have any problem with any $x$-value. For polynomials, the domain is *always* "everything".

Logarithms have restrictions (inputs must be positive); square roots have restrictions (inputs must be non-negative); rational expressions have restrictions (can't divide by zero). Find those restrictions, if present, and remove them from "everything". The domain is everything else.

In your situation:

Can you have a negative inside a square root?

So what needs to be "greater than or equal to zero"?

Also, can you divide by zero?

So what values can $x$ never be?

...but I can't divide the square because a square of -3 is impossible...

I'm sorry, but I don't follow...? Do you perhaps mean that "a square root of -3 is impossible"?

...and if I should make the inequality of all the expression to be great or equal to zero, how to solve this inequality

Solve in the usual way. Set the rational expression "greater than or equal to" zero. Note that the denominator will be positive on two intervals and negative in the middle (thinking of the graph of the associated quadratic function). When multiplied by the negative in the numerator, these intervals will flip. So on which intervals will the rational expression be non-negative?

 

[Steven G]

Domain: $\displaystyle \dfrac{-3}{x^2-5x+4}\geq 0$. Now solve for x

 

[Steven G]

... but I can't divide the square because a square of -3 is impossible, and if I should make the inequality of all the expression to be great or equal to zero, how to solve this inequality

What does it mean to divide the square??? The square of -3 is impossible? I thought that the square of -3 is 9. Why are you even speaking about -3?

By square, do you mean square root? Yes, you can't compute the square root of -3, but no one is asking you to do that! Consider sqrt(-16/-4). You can't compute the sqrt(-16), but again you are not asked to compute the sqrt(-16). You are being asked to compute the sqrt of (-16/-4). Now (-16/-4) = 4. And you CAN compute the sqrt(4)! So in the end the sqrt(-16/-4) = sqrt(4) = 2.

 

[khansaheb]

How can I find the domain of this function?

$\qquad y = \sqrt{\dfrac{-3}{x^2-5x+4}\;}$

I mean, what in the square should be greater or equal to zero, but I can't divide the square because a square of -3 is impossible, and if I should make the inequality of all the expression to be great or equal to zero, how to solve this inequality

The expression under "square-root" sign will be positive, if

x² - 5*x + 4 < 0 →​

​

(x - 4) * (x - 1) < 0​

Now analyze that (↑) condition.

 

[The Highlander]

How can I find the domain of this function?

The expression under "square-root" sign will be positive, if

x² - 5*x + 4 < 0 →​

​

(x - 4) * (x - 1) < 0              and undefined where $\displaystyle x^2 - 5x + 4 = 0$ !​

Now analyze that (↑) condition.

If you graph it you can "see" right away what its domain is (between 1 & 4).

 

[stapel]

The expression under "square-root" sign will be positive, if

$\qquad x^2 - 5x + 4 < 0$

Use what you've learned about graphing quadratic functions:

The quadratic function, $y = x^2 - 5x + 4$, graphs an upward-opening parabola. It will be negative (that is, its value will be less than zero, and its graph will thus be below the $x$-axis) between the two zeroes of $x^2 - 5x + 4 = 0$.

 

[Steven G]

Knowing the zeros of a quadratics and the rule that staplel is talking about in the last post is all you need to know to determine when a quadratic is positive, negative and even zero.

 

[Bruce]

the zeros of a quadratics and the rule that staplel is talking about is all you need to know

Also, if there are no real zeros, we can reason using vertical location of parabola's vertex.

Evaluate vertex y-coordinate using formula: c – b^2/(4a)

 

[Steven G]

Also, if there are no real zeros, we can reason using vertical location of parabola's vertex.

Evaluate vertex y-coordinate using formula: c – b^2/(4a)

Almost.
y= c – b^2/(4a²)

 

[Bruce]

Almost.
y= c – b^2/(4a^2)

Can you show me how you got a^2 on the bottom?

I substituted x = -b/(2a) in ax^2 + bx + c

 

[Dr.Peterson]

Can you show me how you got a^2 on the bottom?

I substituted x = -b/(2a) in ax^2 + bx + c

If $f(x) = ax^2+bx+c$, then $$f\left(-\frac{b}{2a}\right)=a\left(-\frac{b}{2a}\right)^2+b\left(-\frac{b}{2a}\right)+c=a\left(\frac{b^2}{(2a)^2}\right)-\frac{b^2}{2a}+c=\frac{ab^2}{4a^2}-\frac{b^2}{2a}+c=\frac{b^2}{4a}-\frac{2b^2}{4a}+c=c-\frac{b^2}{4a}$$
Looks like you were right. The $a^2$ probably came from not using paper.

A common way to write the final answer is to combine into a single fraction: $$c-\frac{b^2}{4a}=\frac{4ac}{4a}-\frac{b^2}{4a}=\frac{4ac-b^2}{4a}=-\frac{b^2-4ac}{4a}$$
The numerator is the discriminant.

 

[Steven G]

I actually did use paper if I remember correctly. You were right, I was wrong. I will be in the corner for a while.

 

[topsquark]

I actually did use paper if I remember correctly. You were right, I was wrong. I will be in the corner for a while.

Oh, good! I'll have some company after that complex derivative thread. Bring some cards; the guard nicked my deck the last time I was there.

-Dan

 

[Bruce]

−(b^2 − 4ac)/(4a)

The numerator is the discriminant.

I didn't know about that form, thank you. It's helpful when I have the discriminant calculated already.

 

[Steven G]

Oh, good! I'll have some company after that complex derivative thread. Bring some cards; the guard nicked my deck the last time I was there.

-Dan

I'm in for a laugh and will look at that thread now.