Find Real Solutions for rational, radical equations

[snakeyesxlaw]

any help with these would be appreciated..

1) Find all real solutions of the equation:

(5x + 6)/(5x-6) = (2x - 2)/(2x + 2)

2) Find all real solutions of the equation:

2 + ( 4x / (x+3)(x+4) ) = ( 1/(x+3) ) + ( 3/(x+4) )

3) Find all real solutions of the equation:

x - (3-5x)^(1/2) = 0
_______________________
Edited by stapel -- Reason for edit: Removing artifacts of copying from online-test interface

 

[stapel]

1) You cross-multiplied, multiplied out and simplified, got everything together on one side of the "equals" sign, and... then what? Where are you stuck in solving the resulting quadratic equation?

2) Did you start by converting everything to the common denominator, or by multiplying through by the common denominator? Where are you stuck?

3) You converted the "one-half" power to the square root, moved the radical term to the other side of the "equals", squared both sides, and... then what? Where are you stuck?

Please reply showing all of your work and reasoning so far. Thank you!

Eliz.

 

[snakeyesxlaw]

1) You cross-multiplied, multiplied out and simplified, got everything together on one side of the "equals" sign, and... then what? Where are you stuck in solving the resulting quadratic equation?

(5x + 6)/(5x-6) = (2x - 2)/(2x + 2)

okay, i already tried cross-multiplying the first equation and got this;

(5x + 6) * (2x + 2) = (2x - 2) * (5x - 6)
10x^2 + 22x +12 = 10x^2 - 22x + 12
44x = 0

or

10x^2 + 22x +12 = 10x^2 - 22x + 12
2(5x^2 + 11x + 6) = 2(5x^2 - 11x + 6)
2(5x + 6)*(x + 1) = 2(5x-6)*(x-1)
x = -6/5, -1 or x = 6/5, 1

and if you divide the answers, you would get -1
but that was just an educated guess and have never tried to do it that way.

2) Did you start by converting everything to the common denominator, or by multiplying through by the common denominator? Where are you stuck?

2 + ( 4x / (x+3)(x+4) ) = ( 1/(x+3) ) + ( 3/(x+4) )

okay, this is what i did; first multiplied through with the common denominator and got this;

2*(x^2 + 7x + 12) + 4x = (x + 4) + (3x + 9)
2x^2 + 14x + 24 + 4x = 4x + 13
2x^2 + 18x + 24 - 4x - 13 = 0
2x^2 + 14x + 11 = 0

stuck there ^, no perfect square when trying to solve for x

3) You converted the "one-half" power to the square root, moved the radical term to the other side of the "equals", squared both sides, and... then what? Where are you stuck?

x - (3-5x)^(1/2) = 0

okay, this is what i did;

x = (3-5x)^(1/2)
x^2 = (3-5x)
x^2 + 5x -3 = 0

stuck there ^, again not a perfect square to solve for x



any suggestions?

 

[Deleted member 4993]

any help with these would be appreciated..

1) Find all real solutions of the equation:

(5x + 6)/(5x-6) = (2x - 2)/(2x + 2)

2) Find all real solutions of the equation:

2 + ( 4x / (x+3)(x+4) ) = ( 1/(x+3) ) + ( 3/(x+4) )

3) Find all real solutions of the equation:

x - (3-5x)^(1/2) = 0
_______________________
Edited by stapel -- Reason for edit: Removing artifacts of copying from online-test interface

Are these part of test questions? Are you allowed to seek help on these ?

 

[Deleted member 4993]

1) You cross-multiplied, multiplied out and simplified, got everything together on one side of the "equals" sign, and... then what? Where are you stuck in solving the resulting quadratic equation?

(5x + 6)/(5x-6) = (2x - 2)/(2x + 2)

okay, i already tried cross-multiplying the first equation and got this;

(5x + 6) * (2x + 2) = (2x - 2) * (5x - 6)
10x^2 + 22x +12 = 10x^2 - 22x + 12
44x = 0

or

10x^2 + 22x +12 = 10x^2 - 22x + 12
2(5x^2 + 11x + 6) = 2(5x^2 - 11x + 6)
2(5x + 6)*(x + 1) = 2(5x-6)*(x-1)
x = -6/5, -1 or x = 6/5, 1..................How is that??

and if you divide the answers, you would get -1
but that was just an educated guess and have never tried to do it that way.

2) Did you start by converting everything to the common denominator, or by multiplying through by the common denominator? Where are you stuck?

2 + ( 4x / (x+3)(x+4) ) = ( 1/(x+3) ) + ( 3/(x+4) )

okay, this is what i did; first multiplied through with the common denominator and got this;

2*(x^2 + 7x + 12) + 4x = (x + 4) + (3x + 9)
2x^2 + 14x + 24 + 4x = 4x + 13
2x^2 + 18x + 24 - 4x - 13 = 0
2x^2 + 14x + 11 = 0.......................Try solving quadratic equation - should have been covered in algebra I

stuck there ^, no perfect square when trying to solve for x


[quote:1t1x2r8d]
3) You converted the "one-half" power to the square root, moved the radical term to the other side of the "equals", squared both sides, and... then what? Where are you stuck?

x - (3-5x)^(1/2) = 0

okay, this is what i did;

x = (3-5x)^(1/2)
x^2 = (3-5x)
x^2 + 5x -3 = 0.......................Try solving quadratic equation - should have been covered in algebra I


stuck there ^, again not a perfect square to solve for x



any suggestions?[/quote:1t1x2r8d]

 

[Loren]

Seems to me you did the correct steps to arrive at x=0. Did you check that answer? Your second solution involves many steps that only confuse the issue. The steps were okay up to the final resolution.

10x^2 + 22x +12 = 10x^2 - 22x + 12
2(5x^2 + 11x + 6) = 2(5x^2 - 11x + 6)
2(5x + 6)*(x + 1) = 2(5x-6)*(x-1) <<<From here, you can't get to your result.
x = -6/5, -1 or x = 6/5, 1

Did you check each of these four answers?? Let's check x=1.

\(\displaystyle \frac{5(1)+6}{5(1)-6} = \frac{2(1)-2}{2(1)+2}\)

\(\displaystyle \frac{11}{-1} = \frac{0}{0}\) <--- Doesn't check.

 

[stapel]

any help with these would be appreciated....
Edited by stapel -- Reason for edit: Removing artifacts of copying from online-test interface

Are these part of test questions? Are you allowed to seek help on these ?

I don't know if it's an online exam, a quiz, or just electronic homework. It was only clear that the source was a copy-n-paste from an online interface. Since we're trying to help the student learn, we don't need all the white space and the instructions on the order in which to list the answers in the various form fields, etc, etc.

44x = 0

Divide through the complete the solution. Then check your answer.

2(5x + 6)*(x + 1) = 2(5x-6)*(x-1)
x = -6/5, -1 or x = 6/5, 1

Wow.

Whoever told you that this was the way to solve quadratics was very mistaken! :shock: To learn better (such as the Quadratic Formula, factoring, etc, all of which start with the step provided, "getting everything together on one side of the 'equals' sign"), try here or here.

stuck there ^

Follow the links above.

Note: Factoring and solving quadratics should have been covered, in depth, before ever moving on to rational equations. You might want to request a conference with your academic advisor regarding this issue. :idea:

Eliz.

 

[snakeyesxlaw]

much gratitude for the response everyone! this is actually homework through an online system like stapel mentioned earlier.. excuse my carelessness since its been quite awhile since last useing algebra.. totally forgot the quadratic formula for the last two equations and have a bad habit of not excepting 0 as an answer for a problem that emphasizes a real solution..

Problem #1

(5x + 6)/(5x-6) = (2x - 2)/(2x + 2)

okay, already tried cross-multiplying the first equation and got this;

(5x + 6) * (2x + 2) = (2x - 2) * (5x - 6)
10x^2 + 22x +12 = 10x^2 - 22x + 12
44x = 0
x=0/44 = 0

or

{{{{{{ forget i even mentioned this, that approach was ridiculous }}}}}}
10x^2 + 22x +12 = 10x^2 - 22x + 12
2(5x^2 + 11x + 6) = 2(5x^2 - 11x + 6)
2(5x + 6)*(x + 1) = 2(5x-6)*(x-1)
x = -6/5, -1 or x = 6/5, 1
{{{{{{ forget i even mentioned this, that approach was absurd }}}}}}



Problem #2

2 + ( 4x / (x+3)(x+4) ) = ( 1/(x+3) ) + ( 3/(x+4) )

2*(x^2 + 7x + 12) + 4x = (x + 4) + (3x + 9)
2x^2 + 14x + 24 + 4x = 4x + 13
2x^2 + 18x + 24 - 4x - 13 = 0
2x^2 + 14x + 11 = 0

quadratic formula: okay, this is the part on forgetting to apply the quadratic formula..

x = -b +or- square root of b^2 - 4ac divided 2a
x = -24 +or- (14^2 - 4*2*11)/2*2



Problem #3

x - (3-5x)^(1/2) = 0

okay, this is what i did;

x = (3-5x)^(1/2)
x^2 = (3-5x)
x^2 + 5x -3 = 0

quadratic formula: okay, this is the part on forgetting to apply the quadratic formula once again..


just submitted them and got them all right!
many thanks to you all! :wink: