My Math Forum solutions of an equation

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 September 4th, 2011, 09:14 AM #1 Senior Member   Joined: Mar 2011 Posts: 105 Thanks: 0 solutions of an equation 1/x - 1/(x+1) - 3 I solve it and then the solutions are x = (sqrt(21)-3)/6 and x = -(sqrt(21)+3)/6 but my teacher said the number of solutions of an equation depends on the degree of equation why the equation has up to 2 solution but it's not a quadratic equation Can you explain me, please? Thank you so much
 September 4th, 2011, 09:30 AM #2 Senior Member   Joined: Jul 2010 Posts: 103 Thanks: 0 Re: solutions of an equation If you mean: $\frac{1}{x}-\frac{1}{x+1}-3=0$ $\frac{1}{x(x+1)}=3$ $3x^2+3x-1=0$ $x_1 , x_2=\frac{-3\pm\sqrt[]{21}}{6}$ not necessary to put answers to equation and check them instead check the steps that lead you to solve a problem several times.
September 4th, 2011, 09:45 AM   #3
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Re: solutions of an equation

Quote:
 Originally Posted by versmart If you mean: $\frac{1}{x}-\frac{1}{x+1}-3=0$ $\frac{1}{x(x+1)}=3$ $3x^2+3x-1=0$ $x_1 , x_2=\frac{-3\pm\sqrt[]{21}}{6}$ not necessary to put answers to equation and check them instead check the steps that lead you to solve a problem several times.
my teacher said the number of solutions of an equation depends on the degree of equation

why the equation has up to 2 solution

but it's not a quadratic equation

September 4th, 2011, 09:56 AM   #4
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Re: solutions of an equation

Quote:
 Originally Posted by versmart If you mean: $\frac{1}{x}-\frac{1}{x+1}-3=0$ $\frac{1}{x(x+1)}=3$ $3x^2+3x-1=0$ $x_1 , x_2=\frac{-3\pm\sqrt[]{21}}{6}$ not necessary to put answers to equation and check them instead check the steps that lead you to solve a problem several times.
I was thinking like you, not necessary to check them

but for example : abs(x) = x^2 + x - 3
If you don't check them, there are 4 solutions, but if you check, only 2 of them are solution

So, can you or someone else tell me when I should check the solution?

Thank you so much

September 4th, 2011, 01:17 PM   #5
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Re: solutions of an equation

Quote:
 my teacher said the number of solutions of an equation depends on the degree of equation why the equation has up to 2 solution but it's not a quadratic equation
"The maximum number of solutions to an equation with degree n is n"
for example $Ax^n+Bx^{n-1}+Cx^{n-2}+...+z$ could have n solutions or n-m solutions while m<n
special case :
quadratic equation = $Ax^2+Bx+C=0$

0-reform the equation to the form of $Ax^2+Bx+C=0$

1- find $\Delta=B^2-4AC$

2- if $\Delta > 0$ so there are 2 roots $x_1 , x_2= \frac{-B \pm \sqrt[]{\Delta}}{2A}$
if $\Delta= 0$ so there are 2 roots and equal $x_1=x_2 = \frac{-B }{2A}$
if $\Delta < 0$ there is no real root but complex $x_1 , x_2= \frac{-B \pm i\sqrt[]{-\Delta}}{2A}$

Quote:
 but for example : abs(x) = x^2 + x - 3 If you don't check them, there are 4 solutions, but if you check, only 2 of them are solution
how you found 4 solutions to that equation?
$x^2+x-3=0$there are only 2 roots and it's not possible to have more roots:
$x_1 , x_2= \frac{-1\pm sqrt[]{\sqrt[]{13}}}{2}$

you should not make another problem for yourself but if you like to check I suggest you this technique:
let $x_1 , x_2$ be the roots of $Ax^2+Bx+C=0$
always and always:
$x_1*x_2=\frac{C}{A}$
$x_1+x_2=\frac{-B}{A}$

I use this if I want to check:
$|x_1-x_2|=\frac{\sqrt[]{\Delta}}{|A|}$

September 4th, 2011, 02:27 PM   #6
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Re: solutions of an equation

Quote:
 but it's not a quadratic equation
Then what do you call this?:
Quote:
 Originally Posted by daivinhtran $3x^2+3x-1=0$
Quote:
 not necessary to put answers to equation and check them instead check the steps that lead you to solve a problem several times.
Yes, but you can still flub up, or perhaps just not know what you're doing, so checking the answers is certainly never a bad idea. So I don't see any "instead" here. It's actually much easier to miss mistakes when you're checking over your work because you tend to see what you think you should, and it's terrifically easy to miss signs and such. In fact, the common mistakes people make can be just the kind of mistake you'd miss (again) when you check it over.

So while it doesn't hurt to check back over your steps, it's also a terrifically good idea to check your answers as well, which I think gives me more comfort that what I did was correct.

September 4th, 2011, 11:21 PM   #7
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Re: solutions of an equation

Quote:
 So while it doesn't hurt to check back over your steps, it's also a terrifically good idea to check your answers as well, which I think gives me more comfort that what I did was correct.
if you pay attention i gave him a simpler method of checking his answers if the equation is second degree:

Quote:
 I use this if I want to check: $|x_1-x_2|=\frac{\sqrt[]{\Delta}}{|A|}$
it is faster than putting 2 answers 2 times into and equation which most of the times need calculator!

September 4th, 2011, 11:38 PM   #8
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Re: solutions of an equation

Quote:
Originally Posted by versmart
Quote:
 it is faster than putting 2 answers 2 times into and equation which most of the times need calculator!
And most of the time I have a calculator. I was simply making a point that there's something to be said for checking answers and that there's certainly nothing wrong with it. Not to mention that I was answering your comment that you should "instead check the steps that led you to solve the problem."

September 7th, 2011, 02:14 PM   #9
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Re: solutions of an equation

Quote:
Originally Posted by versmart
Quote:
 [quote:1fexck7o]but for example : abs(x) = x^2 + x - 3 If you don't check them, there are 4 solutions, but if you check, only 2 of them are solution
how you found 4 solutions to that equation?
$x^2+x-3=0$there are only 2 roots and it's not possible to have more roots:
$x_1 , x_2= \frac{-1\pm sqrt[]{\sqrt[]{13}}}{2}$

[/quote:1fexck7o]

honestly, I have to say you didn't read the question carefully
my question is abs(x) = x^2 + x - 3
it's ABS(x) = |x|

You can't find the correct answer without checking....

September 7th, 2011, 05:14 PM   #10
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Re: solutions of an equation

Quote:
 Originally Posted by daivinhtran honestly, I have to say you didn't read the question carefully my question is abs(x) = x^2 + x - 3 it's ABS(x) = |x| You can't find the correct answer without checking....
Hello daivinhtran,

Your example may not have only two solutions because it is not a polynomial equation. What your teacher said only applies to polynomials.

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