Solving Equation Containing a Radical (Square Root)

[holmes1172]

I feel I'm on the right path here, just tripped somewhere...help?

x-√3x+18=0
x-√3x+18
x^2 √3x+18^2
x^2=18+3x
x^2+3x-18=0
(x+3)(x+6)=0
X+3+0 or x+6=0
Checking
X=-3
-3-√3(-3)+18=0
-3—√-9+18=0
-3—9=0 False
Checking
-6-√3(6)+18=0
-6-√-18+18=0

 

[pka]

I feel I'm on the right path here, just tripped somewhere...help?

x-√3x+18=0

First lets see what is the right question.

Is it \(\displaystyle x-\sqrt{3}\;x+18~=0\text{ or }x-\sqrt{3x}+18~=0~?\)

If it is the second then you should have written x-√{3x}+18=0.

 

[holmes1172]

First lets see what is the right question.

Is it \(\displaystyle x-\sqrt{3}\;x+18~=0\text{ or }x-\sqrt{3x}+18~=0~?\)

If it is the second then you should have written x-√{3x}+18=0.

It is the second...

 

[pka]

It is the second...

Then let \(\displaystyle x=y^2\) solve \(\displaystyle y^2-\sqrt{3}|y|+18=0\)

 

[holmes1172]

Then let \(\displaystyle x=y^2\) solve \(\displaystyle y^2-\sqrt{3}|y|+18=0\)

x - √ (3x + 18) = 0

Let x = y^2

y^2 - √ (3 |y| + 18) = 0

Dividing ‘y’ out I get:

y - √ (3 + 18) = 0

Now, from here would I go ahead and add the 3+18. My example in my textbook has me factoring. Would it maybe now be:

y + 3 + 18 = 0?

 

[Nazariy]

Then let \(\displaystyle x=y^2\) solve \(\displaystyle y^2-\sqrt{3}|y|+18=0\)

Does this even have solutions?:roll:

 

[lookagain]

???

holmes1172, you must have copied the problem incorrectly. If you were given a radical equation with
a square root, and if the solution(s) was/were supposed to come out rational (my supposition),
then I think the problem you were actually given was either


\(\displaystyle x - \sqrt{3x} - 18 = 0 \ \ \ \ or\)


\(\displaystyle x - \sqrt{3x} = 18\)



These are equivalent equations and have an integer for their solutions.

 

[pka]

Does this even have solutions?

OH! Are you sure of that. Look at this calculation.

 

[Nazariy]

OH! Are you sure of that. Look at this calculation.

Oh dear, complex numbers, my competency ends here, for now!

 

[holmes1172]

holmes1172, you must have copied the problem incorrectly. If you were given a radical equation with
a square root, and if the solution(s) was/were supposed to come out rational (my supposition),
then I think the problem you were actually given was either


\(\displaystyle x - \sqrt{3x} - 18 = 0 \ \ \ \ or\)


\(\displaystyle x - \sqrt{3x} = 18\)



These are equivalent equations and have an integer for their solutions.

I went back and checked, and the problem I was given was:

x - √ (3x + 18) = 0

All the instructions say are to "Solve each equation" and see examples 4-6 which are Solving an Equation Containing a Radical (Square Root), Solving an Equation Containing Two Radicals, and Solving an Equation Containing a Radical (Cube Root). I thought it appeared more like the Square Root solution, so I was going with that. Maybe I should've chosen one of the other two examples, you think?

 

[Quaid]

I went back and checked, and the problem I was given was:

x - √ (3x + 18) = 0

I do not understand. You already checked the problem.

First lets see what is the right question.

Is it \(\displaystyle x-\sqrt{3}\;x+18~=0\text{ or }x-\sqrt{3x}+18~=0~?\)

It is the second

This is unfortunate.