Rearrange Formula: 2/(x-2) + 2/(x+2) = 1/2 to Ax2+ Bx + C = 0

[Natasha1]

1. Rearrange the following equation
2/(x-2) + 2/(x+2) = 1/2
into the form Ax2+ Bx + C = 0

I did it and got -x2 + 8x + 4 = 0

Am I correct please?

 

[berkeman]

1. Rearrange the following equation
2/(x-2) + 2/(x+2) = 1/2
into the form Ax2+ Bx + C = 0

I did it and got -x2 + 8x + 4 = 0

Am I correct please?

Could you please show your step-by-step work? That will make it easier for us to check it.

 

[FactChecker]

Looks good to me. You should usually show your work in step-by-step fashion. Trying to do too much in your head is slow and error-prone.

 

[Natasha1]

See picture attached for working out

 

[Ray Vickson]

1. Rearrange the following equation
2/(x-2) + 2/(x+2) = 1/2
into the form Ax2+ Bx + C = 0

I did it and got -x2 + 8x + 4 = 0

Am I correct please?

No, your final equation in incorrect. Start again, and this time do it the easy way: re-write $2/(x-2) + 2/(x+2)$ by putting both terms over a common denominator. I really could not figure out what you were trying to do in your original working, but whatever it was produced errors.

 

[Natasha1]

Where am I going wrong... I need to rearrange the following equation 2/(x-2) + 2/(x+2) = 1/2 into the form Ax2+ Bx + C = 0

Here's my work:

2/(x-2) + 2/(x+2) = 1/2

2 + 2(x-2)/(x+2) - (x-2)/2 = 0
2(x+2) + 2(x-2) - (x-2)(x+2)/2 = 0
4(x+2)+4(x-2)-(x-2)(x+2) = 0
4x+8+4x-8-(x2 - 2x+2x-4) = 0
8x-x2+4 = 0
-x2 +8x+4 = 0

 

[FactChecker]

I think that your answer is correct ... with two caveats.
1) Always keep track of values of x that are invalid: x=2 or x=-2 cause a divide by 0 in the original equation. So always add the constraints x≠2; x≠-2. (Click on the 'Σ' button at the top of the entry window to get the '≠' symbol, among others.)
2) Use clear notation for the exponent 2. If you do not have superscripts, type x^2. (On the web site, click on the 'x²' button at the top of the entry window to get superscripts.)

So your final equation should look like:
-x² +8x+4 = 0; x ≠ +2; x ≠ -2

 

[Ray Vickson]

Where am I going wrong... I need to rearrange the following equation 2/(x-2) + 2/(x+2) = 1/2 into the form Ax2+ Bx + C = 0

Here's my work:

2/(x-2) + 2/(x+2) = 1/2

2 + 2(x-2)/(x+2) - (x-2)/2 = 0
2(x+2) + 2(x-2) - (x-2)(x+2)/2 = 0
4(x+2)+4(x-2)-(x-2)(x+2) = 0
4x+8+4x-8-(x2 - 2x+2x-4) = 0
8x-x2+4 = 0
-x2 +8x+4 = 0

Sorry: I missed the minus sign in front of your x^2 term, so $x^2 -8x - 4=0$ IS correct. You wrote it as $-x^2 + 8x + 4 = 0$, but the usual way of writing polynomial equations is to have the coefficient of the highest power of $x$ being positive (so with $+x^2$ instead of $-x^2$).

 

[Natasha1]

I think that your answer is correct ... with two caveats.
1) Always keep track of values of x that are invalid: x=2 or x=-2 cause a divide by 0 in the original equation. So always add the constraints x≠2; x≠-2. (Click on the 'Σ' button at the top of the entry window to get the '≠' symbol, among others.)
2) Use clear notation for the exponent 2. If you do not have superscripts, type x^2. (On the web site, click on the 'x²' button at the top of the entry window to get superscripts.)

So your final equation should look like:
-x² +8x+4 = 0; x ≠ +2; x ≠ -2

I'm sorry thank you for the advice

 

[Natasha1]

Sorry: I missed the minus sign in front of your x^2 term, so $x^2 -8x - 4=0$ IS correct. You wrote it as $-x^2 + 8x + 4 = 0$, but the usual way of writing polynomial equations is to have the coefficient of the highest power of $x$ being positive (so with $+x^2$ instead of $-x^2$).

Why is it then I get two different solution for both equations -x^2 + 8x +4 = 0 and x^2 -8x - 4 = 0 ?

 

[Ray Vickson]

Why is it then I get two different solution for both equations -x^2 + 8x +4 = 0 and x^2 -8x - 4 = 0 ?

They are not two different equations! If $x^2 -8x-4=0$ then also $0 = -0 = -(8x^2 -8x-4) = -x^2 + 8x + 4$.

OK, I guess you could say they are different, just as $5x^2 - 40 x - 20=0$ looks different; but they are all just simple multiples of each other, and they all have exactly the same solutions.

 

[SammyS]

They are not two different equations! If $x^2 -8x-4=0$ then also $0 = -0 = -(8x^2 -8x-4) = -x^2 + 8x + 4$.

OK, I guess you could say they are different, just as $5x^2 - 40 x - 20=0$ looks different; but they are all just simple multiples of each other, and they all have exactly the same solutions.

@Natasha1 ,

Thus they are called equivalent equations .