My Math Forum Inequality solve

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 December 22nd, 2010, 04:46 PM #1 Member   Joined: Dec 2010 From: Viet Nam and US Posts: 36 Thanks: 0 Inequality solve Hi guys I'm new, I need help in this topic Solve: $6x^2+9x+10x\sqrt{3x+2}-1\le 0$ [color=#008000] (1)[/color] I'm trying to use another variable to replace x (sorry about my English, I'm Asian, I just love math ). So I do: Let $t=\sqrt{3x+2}\leftrightarrow 3t^2=9x+6$ [color=#008000] (2)[/color] [color=#008000] (1)(2)[/color]$\leftrightarrow 6x^2+3t^2-10t-7\le 0$ And I'm stuck there
 December 22nd, 2010, 05:38 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,469 Thanks: 2038 Where did "-10t" spring from? You have 3x = t² - 2, so multiply the original equation by 3 and then substitute for 3x throughout and find the non-negative values of t that satisfy the inequality, etc.
December 22nd, 2010, 05:56 PM   #3
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Quote:
 Originally Posted by skipjack Where did "-10t" spring from? You have 3x = t² - 2, so multiply the original equation by 3 and then substitute for 3x throughout and find the non-negative values of t that satisfy the inequality, etc.
I'm not sure what are you trying to help me out, could you do it for me?

About where "-10t" comes from, I mistyped, it must be "-10xt" let me explain:
We have $t=\sqrt{3x+2}$ and $3t^2=9x+6$
The original inequality can be rewritten:
$6x^2+[9x+(6-7)]+10x\sqrt{3x+2}\le 0$ or $tx^2+3t^2+10xt-7\le 0$
And that's how I came up, I need another better solution I guess

 December 22nd, 2010, 06:10 PM #4 Global Moderator   Joined: Dec 2006 Posts: 20,469 Thanks: 2038 That's precisely what I gave you - multiply the original equation by 3, etc. You'll find it easy if you try it. Alternatively, plot the graph of the left-hand side of the original equation.
December 22nd, 2010, 07:33 PM   #5
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Re:

Quote:
 Originally Posted by skipjack That's precisely what I gave you - multiply the original equation by 3, etc. You'll find it easy if you try it. Alternatively, plot the graph of the left-hand side of the original equation.
I've done what you told (I don't know whether I understand it correctly or not), and thanks to you, I have this:

$6x^2+9x+10x\sqrt{3x+2}-1\le 0$

Multiply both sides by 3, substitute and simplify:
$\Leftrightarrow\begin{cases}18x^2+27x+30x\sqrt{3x+ 2}-3\le 0\\t=\sqrt{3x+2}\Leftrightarrow t^2=3x+2\Leftrightarrow 3x=t^2-2\end{cases}$
$2t^4+10t^3+t^2-20t+10\le 0$

Factor out (I don't know how to explain, just see the line below) the result, we have:
$\Leftrightarrow(t-\sqrt{2})(t+\sqrt{2})(2t^2+10t+5)\le 0$

Listed the solutions orderly, we have:

$\begin{vmatrix}
-\infty & \frac{-10-2\sqrt{15}}{4} & -\sqrt{2} & \frac{-10+2\sqrt{15}}{4} & \sqrt{2} &+\infty \\
+ & +(0)- & - & - & - & -\\
- & - & -(0)+ & + & + & +\\
- & - & - & -(0)+ & + & +\\
- & - & - & - & -(0)+ & +
\end{vmatrix}$

So in order to make the equation satisfy, we have to pick:

$t\in (-\infty;\frac{-10-2\sqrt{15}}{4}]\cup [-\sqrt{2}; \frac{-10+2\sqrt{15}}{4}]\cup [\sqrt{2};+\infty]$

But that is the way we pick t, how about x? The main reason is to solve x. I think I mislead somewhere!

December 22nd, 2010, 07:48 PM   #6
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Quote:
 Originally Posted by xx3004 Multiply both sides by 3, substitute and simplify: $2t^4+10t^3+t^2-20t+10\le 0$
Your simplification was incorrect (the constant at the end should not be 10), and your factorization of the polynomial you obtained was also incorrect; the correct polynomial factorizes fairly easily. When you solve, remember that only non-negative values of t are of interest.

December 22nd, 2010, 08:09 PM   #7
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Re:

Quote:
Originally Posted by skipjack
Quote:
 Originally Posted by xx3004 Multiply both sides by 3, substitute and simplify: $2t^4+10t^3+t^2-20t+10\le 0$
Your simplification was incorrect (the constant at the end should not be 10), and your factorization of the polynomial you obtained was also incorrect; the correct polynomial factorizes fairly easily. When you solve, remember that only non-negative values of t are of interest.
I see, my fault, I'm so carelessly! Now let's start over

$6x^2+9x+10x\sqrt{3x+2}-1\le 0$

Multiply both sides by 3, substitute and simplify:
$\Leftrightarrow\begin{cases}18x^2+27x+30x\sqrt{3x+ 2}-3\le 0\\t=\sqrt{3x+2}\Leftrightarrow t^2=3x+2\Leftrightarrow 3x=t^2-2\end{cases}$
$2t^4+10t^3+t^2-20t-13\le 0$

Factor out and we have:
$\Leftrightarrow(x+1)^2(2t^2+10t+5)\le 0$

Listed the solutions orderly, we have:

$\begin{vmatrix}
-\infty & \frac{-3-\sqrt{35}}{2} & {-1} & \frac{-3+\sqrt{35}}{2} & +\infty\\
+ & +(0)- & - & - & -\\
+ & + & +(0)- & - & -\\
- & - & - & -(0)+ & +
\end{vmatrix}$

So in order to make the equation satisfy, we have to pick:

$t\in (-\infty;\frac{-3-\sqrt{35}}{4}]\cup [-1; \frac{-3+\sqrt{35}}{2}]$

I think it's ok now, but how about x? That solution now works for t only. I'm not sure what you mean by "only non-negative values of t are of interest" (language limit I'm sorry).

 December 22nd, 2010, 08:52 PM #8 Global Moderator   Joined: Dec 2006 Posts: 20,469 Thanks: 2038 Your (x + 1)² should be (t + 1)², but can be omitted (just solve 2t² + 10t + 5 ? 0). You solved incorrectly. You need t ? 0, so 0 ? t ? (-3 + ?35)/2. Hence 0 ? t² ? (22 - 3?35)/2, i.e., 0 ? 3x + 2 ? (22 - 3?35)/2, and so -2/3 ? x ? (6 - ?35)/2.
December 22nd, 2010, 09:00 PM   #9
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Re:

Quote:
 Originally Posted by skipjack Your (x + 1)² should be (t + 1)², but can be omitted. You solved incorrectly. You need t ? 0, so 0 ? t ? (-3 + ?35)/2. Hence 0 ? t² ? (22 - 3?35)/2, i.e., 0 ? 3x + 2 ? (22 - 3?35)/2, and so -2/3 ? x ? (6 - ?35)/2.
I gotcha, thank you very much Skipjack!

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