# Polynomial, Radical Equations & Inequalities. Help needed by Jan. 9,2007!

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• Jan 4th 2007, 12:36 AM
earboth
Hello,

to solve equations which contain square roots you have to isolate the square root on one side of the equation and then square the complete equation. But be careful: This transformation will not give an equivalent equation. So you have to check your answer:

$\sqrt{2x-3}+6=10 \Longleftrightarrow \sqrt{2x-3}=4 \Longrightarrow 2x-3=16 \Longrightarrow 2x=19 \Longrightarrow x=\frac{19}{2}$

Now plug in this number into the original equation:

$\sqrt{2 \cdot \frac{19}{2}-3}+6=10 \Longleftrightarrow \sqrt{16}+6=10 \Longleftrightarrow 10=10$ which is obviously true. Therefore x=9.5 is the solution of the equation.

EB
• Jan 4th 2007, 12:39 AM
earboth
Hello,

#18:

$4x^2+36=0 \Longleftrightarrow 4x^2=-36 \Longleftrightarrow x^2=-9$

Now calculate the square root of both sides of this equation:

$x=3i\ \vee \ x=-3i$

EB
• Jan 4th 2007, 05:49 AM
ThePerfectHacker
Quote:

Originally Posted by earboth

$\sqrt{-1}=i$

It is more appropriate to define,
$i^2=-1$.
• Jan 4th 2007, 07:59 AM
earboth
Hello,

it's me again. (I started this post but suddenly it has vanished without a trace...:confused: )

To #13 and #14: I don't know the difference between long division and synthetic division (that has nothing to do with my limited kowledge of Math but with the incompleteness of my dictionary!) I'll do both problems the same way - and hope that you can use my reply nevertheless:

to #13:
Code:

 (3x^3 + x^2 + 4x - 5)÷(3x-2) = x^2+x+2, remainder (-1)  -(3x^3 -2x^2) --------------         3x^2 + 4x       -(3x^2 - 2x)           ------------                 6x - 5               -(6x - 4)               --------                     -1
I've learned that the result is:

$(3x^3+x^2+4x-5)\div (3x-2)=x^2+x+2-\frac{1}{3x-2}$

to #14:
Code:

  (x^4 - 3x^3 - 2x^2 + 1)÷(x-3) = x^3-2x-6, remainder -17 -(x^4 - 3x^3) --------------                 -2x^2 + 1               -(-2x^2+6x)                 -------------                       -6x + 1                     -(-6x + 18)                     -----------                             -17
That means:

$(x^4-3x^3-2x^2+1) \div (x-3)=x^3-2x-6-\frac{17}{x-3}$

EB
• Jan 4th 2007, 03:54 PM
Mulya66
Thanks so very much! The long division/synthetic division problems are the one's that we've always had trouble with. The synthetic division is very new to me because I've never seen it before, but thank you for your time and help!

Quote:

Originally Posted by earboth
Hello,

it's me again. (I started this post but suddenly it has vanished without a trace...:confused: )

To #13 and #14: I don't know the difference between long division and synthetic division (that has nothing to do with my limited kowledge of Math but with the incompleteness of my dictionary!) I'll do both problems the same way - and hope that you can use my reply nevertheless:

to #13:
Code:

 (3x^3 + x^2 + 4x - 5)÷(3x-2) = x^2+x+2, remainder (-1)  -(3x^3 -2x^2) --------------         3x^2 + 4x       -(3x^2 - 2x)           ------------                 6x - 5               -(6x - 4)               --------                     -1
I've learned that the result is:

$(3x^3+x^2+4x-5)\div (3x-2)=x^2+x+2-\frac{1}{3x-2}$

to #14:
Code:

  (x^4 - 3x^3 - 2x^2 + 1)÷(x-3) = x^3-2x-6, remainder -17 -(x^4 - 3x^3) --------------                 -2x^2 + 1               -(-2x^2+6x)                 -------------                       -6x + 1                     -(-6x + 18)                     -----------                             -17
That means:

$(x^4-3x^3-2x^2+1) \div (x-3)=x^3-2x-6-\frac{17}{x-3}$

EB

• Jan 4th 2007, 09:33 PM
earboth
Quote:

Originally Posted by Mulya66
Thanks so very much! The long division/synthetic division problems are the one's that we've always had trouble with. The synthetic division is very new to me because I've never seen it before, but thank you for your time and help!

Hello,

I finally found out that I know the so-called synthetic division under the name "Horner's scheme". To get further information have a look here:
Synthetic Division

EB
• Jan 5th 2007, 03:19 PM
Mulya66
Quote:

Originally Posted by earboth
Hello,

I finally found out that I know the so-called synthetic division under the name "Horner's scheme". To get further information have a look here:
Synthetic Division

EB

Hi,

Unfortunately, I've already been to Purple Math and we tried their way and the Algebra book's way as well. That's why I said the synthetic/long division problems were the hardest. :(

Thanks again. :)
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