MHB How do I solve this inequality with positive coefficients?

[mathdad]

Solve (ax + b)/(root{x}) > 2*root{ab}, where a > 0, b > 0.

Can someone provide the steps or at least get me started?

 

[skeeter]

Solve (ax + b)/(root{x}) > 2*root{ab}, where a > 0, b > 0.

Can someone provide the steps or at least get me started?

$\dfrac{ax+b}{\sqrt{x}} > 2\sqrt{ab}$

since $\sqrt{x} > 0$ ...

$ax+b > 2\sqrt{abx}$

$(ax)^2 + 2abx + b^2 > 4abx$

$(ax)^2 - 2abx + b^2 > 0$

$(ax-b)^2 > 0$

finish it ...

 

[mathdad]

I am solving for x, right?

Is this a quadratic inequality?

 

[skeeter]

I am solving for x, right?

yes

Is this a quadratic inequality?

looks like one to me ...

 

[mathdad]

Cool. I will solve it early this evening. I am working now.

 

[skeeter]

solution set for the original inequality ...

$x \in \left(0,\dfrac{b}{a}\right) \cup \left(\dfrac{b}{a},\infty\right)$

 

[mathdad]

(ax - b)^2 > 0

sqrt{(ax - b)^2} = sqrt{0}

ax - b = 0

ax = b

x = b/a<------------(b/a)----------->Is this the correct number line for testing each interval in this case?

 

[skeeter]

<------------(b/a)----------->Is this the correct number line for testing each interval in this case?

no. should look like this ...

0(------------(b/a)----------->

why?

 

[mathdad]

I was thinking of testing each interval leading to the solution. How did you determine the solution you provided?

 

[skeeter]

I was thinking of testing each interval leading to the solution. How did you determine the solution you provided?

I tested intervals ... look again at the original inequality. What is the first thing you can say about the domain of $x$?

 

[mathdad]

I want to learn how to test the interval of this strange question. I see that you provided the solution. The domain of x in given problem must be [1, infinity).

 

[skeeter]

The domain of x in given problem must be [1, infinity).

no.

the denominator on the left side of the original inequality is $\sqrt{x}$, which has to be strictly greater than zero $\implies x > 0$

once you're to this point from doing the algebra on the original inequality ...

$(ax-b)^2 > 0$

... any non-zero value squared is positive, so $ax-b \ne 0 \implies x \ne \dfrac{b}{a}$. Couple that fact with the previous determination that $x > 0$ leads to the solution set in post #6.

 

[mathdad]

Skeeter,

Allow me to say a few things in the form of an outline.

1. I am not a student in a formal classroom. My college days ended in 1994.

2. I am 52 years old.

3. I love math. I am trying to review precalculus one chapter at a time through self-study and this website.

4. I need math to be explained in basic terms. To answer my questions in a textbook-like form is no different than reading the David Cohen precalculus book on my own.

5. You must be patient with me as I review precalculus. I took this course in the Spring 1993 semester and got an A minus. Not bad for an elective course. So, please be patient as I endeavor to review material learned long ago.

 

[skeeter]

Understand this is not news to me ... I've read this same series of statements regarding your situation before.

 

[mathdad]

Cool. So, this means you will try to be patient with me here. As you know, I try to show my work to posted questions.