surd equation assistance: sqrt(9x^2+1) + sqrt(12x/5+4) = 3(sqrt(x^2+8x/5+1)

[grant]

Hi all,

I think that this question might fit in this forum, although it may be more of a beginner question, i'm not sure.

Anyhow, I am trying to solve for x in the following:

sqrt(9x^2+1) + sqrt(12x/5+4) = 3(sqrt(x^2+8x/5+1)

What i have done so far is:

square both sides: 9x^2+1+12x/5+4 + 2(sqrt((9x^2+1)(12x/5+4)) = 9(x^2+8x/5+1)

Simplified and moved the non-sqrt expressions to the right hand side, and multiplied out the brackets on the right hand side:
2(sqrt((9x^2+1)(12x/5+4))= 9x^2-9x^2+72x/5-12x/5+1-5

Simplifying further: 2(sqrt((9x^2+1)(12x/5+4))=60x/5-4

and further: 2(sqrt((9x^2+1)(12x/5+4))=12x-4

dividing both sides by 2: sqrt((9x^2+1)(12x/5+4))=6x-2

Squaring both sides again: (9x^2+1)(12x/5+4)=36x^2-24x+4

Here is where i have trouble, because as soon as i multiply out the bracket on the left side I get an x^3 which first of all, i have not learnt how to do yet, and secondly, I think the problem can be solved without factorising a cubed expression.
The text book has used an example where the initial part of a quadratic expression is set as 'y' (e.g 2x^+3x+5, let y= 2x^2+3x) but I cannot see any opportunities to do that here as I cannot eliminate all the x s with one expression (which i would then set to y, solve for y, and then solve for x). Anyway, to continue:

multiply out the expression:108x^3/5+36x^2+12x/5+4=36x^2-24x+4

the 4 and 36x^2 s cancel out leaving: 108x^3/5+12x/5=24x

here I can divide both sides by x:108x^2/5+12/5-24 = 0

then multiply by 5: 108x^2-108 = 0

I'm not really sure how to solve for x from this point.

I hope my layout is clear! this problem is gnawing away at me and making me feel quite stupid!! :/

Any help would be appreciated

 

[Deleted member 4993]

Hi all,

I think that this question might fit in this forum, although it may be more of a beginner question, i'm not sure.

Anyhow, I am trying to solve for x in the following:

sqrt(9x^2+1) + sqrt(12x/5+4) = 3(sqrt(x^2+8x/5+1)

What i have done so far is:

square both sides: 9x^2+1+12x/5+4 + 2(sqrt((9x^2+1)(12x/5+4)) = 9(x^2+8x/5+1)

Simplified and moved the non-sqrt expressions to the right hand side, and multiplied out the brackets on the right hand side:
2(sqrt((9x^2+1)(12x/5+4))= 9x^2-9x^2+72x/5-12x/5+1-5

Simplifying further: 2(sqrt((9x^2+1)(12x/5+4))=60x/5-4

and further: 2(sqrt((9x^2+1)(12x/5+4))=12x-4

dividing both sides by 2: sqrt((9x^2+1)(12x/5+4))=6x-2

Squaring both sides again: (9x^2+1)(12x/5+4)=36x^2-24x+4

Here is where i have trouble, because as soon as i multiply out the bracket on the left side I get an x^3 which first of all, i have not learnt how to do yet, and secondly, I think the problem can be solved without factorising a cubed expression.
The text book has used an example where the initial part of a quadratic expression is set as 'y' (e.g 2x^+3x+5, let y= 2x^2+3x) but I cannot see any opportunities to do that here as I cannot eliminate all the x s with one expression (which i would then set to y, solve for y, and then solve for x). Anyway, to continue:

multiply out the expression:108x^3/5+36x^2+12x/5+4=36x^2-24x+4

the 4 and 36x^2 s cancel out leaving: 108x^3/5+12x/5=24x

here I can divide both sides by x:108x^2/5+12/5-24 = 0 .............. we cannot do that because x could equal to zero.

then multiply by 5: 108x^2-108 = 0

I'm not really sure how to solve for x from this point.

I hope my layout is clear! this problem is gnawing away at me and making me feel quite stupid!! :/

Any help would be appreciated

Assuming your arithmetic is correct, we have

108x^3/5 + 12x/5 - 24x = 0

x(108 x^2 - 108) = 0

108 * x (x^2 - 1) = 0

108 * x * (x - 1) * (x + 1) = 0

What are the possible values of 'x' - that will make the above equation true?

Also evaluate your original function at these values of 'x's and confirm our answers.

 

[grant]

Thank you very much.

that answer wasn't what i expected (although very much appreciated), I thought I had made a fundamental error right at the beginning, or that a completely new approach at the start (e.g. reorganising the equation using a variable to replace the expressions containing x).

I checked through and it seems that 1 and 0 were the correct answers.

thanks again.