Equation with sqrt of x

[Arne]

Sorry but i really dont understand how to write this properly from android.

I have an equation.
3sqrt(x) + sqrt(3 + x) = sqrt(x)

I know you can subtract with 3sqrt(x) on both sides now, but i want to quadrate first.

And i get
2(sqrt(9x(3 + x)) = -9x -3

And
108x +36x² = 81x² + 54x + 9

And (54 +/- 36) ÷ 90

X₁ = 1
X₂ = 0.2
Which both are false.

The problem is that this is an old exam question from last year and they do the subtracting forst then quadrating and end up with only x = 1 which is false. But i get 2 false roots. Normally i would think the solution is flawed.. But this being on last years exam i feel it should be right.

So am i doing something wrong? Or is the solution actually flawed?

 

[stapel]

Sorry but i really dont understand how to write this properly from android.

I have an equation.
3sqrt(x) + sqrt(3 + x) = sqrt(x)

That's exactly how to "write this properly". Thank you! :cool:

I know you can subtract with 3sqrt(x) on both sides now, but i want to quadrate first.

And i get
2(sqrt(9x(3 + x)) = -9x -3

I'm sorry, but what does it mean to "quadrate" a radical equation? How did doing so result in the last line above?

When you reply, please show all of your steps. Thank you!

 

[ksdhart]

The main reason you'd want to do the subtraction/grouping like terms first, is to make the math a bit easier to do in your head. Asqrt(a)=Bsqrt(b) is a lot easier to work with than Asqrt(a) + Bsqrt(b) = Csqrt(c). That said, squaring the function first is a valid method of solving the problem. The difficulty here is when you square both sides of an equation, you do sometimes end up with false roots. In this case, both roots are false, so there are no real solutions. Teachers have been known to assign unsolvable equations, just to keep you on your toes and make sure you're paying attention.

 

[Arne]

The whole point

Was to not rearange the equation. By qaudrating i mean squaring both sides of the equation twice to get rid of all squareroots.

So i set
(3sqrt(x) + sqrt(3 + x))² = sqrt(x)²

Which leads to 9x + 2(3sqrt(x) * sqrt(x + 3)) + 3 + x = x

And i swap it around.
-9x -3 = 2(3sqrt(x) * sqrt(3 + x))

And sqrt(a) * sqrt(b) = sqrt (a * b), and a * sqrt(b) = sqrt(a²b) so..

-9x -3 = 2(sqrt(9x(3 + x))

And i sqare the second time getting.

(-9x-3)(-9x-3) = 4(9x(x + 3)

Then i put the result through the formula x = ( -b +/- sqrt(b² - 4ac)) ÷ 2a

And get x = (54 +/- 36) ÷ 90

And the x = 1 root is okay, i know that one is right because of the solution and i know its false. But my question is about x = 0.2 since that root doesnt exist in the solution since they did the approach by rearanging like Denis suggests, i also did that first but it hit me that i might get another answer by doing it the long hard way..

And they teach us that you hvae to find all possible roots and try them if they exist. But since this is an old exam im wondering if im doing something wrong, or if the solution is wrong.

 

[Deleted member 4993]

Was to not rearange the equation. By qaudrating i mean squaring both sides of the equation twice to get rid of all squareroots.

So i set
(3sqrt(x) + sqrt(3 + x))² = sqrt(x)²

Which leads to 9x + 2(3sqrt(x) * sqrt(x + 3)) + 3 + x = x

And i swap it around. (Now you are trying to re-arrange - Hummmmm)
-9x -3 = 2(3sqrt(x) * sqrt(3 + x))

And sqrt(a) * sqrt(b) = sqrt (a * b), and a * sqrt(b) = sqrt(a²b) so..

-9x -3 = 2(sqrt(9x(3 + x))

And i sqare the second time getting.

(-9x-3)(-9x-3) = 4(9x(x + 3)

Then i put the result through the formula x = ( -b +/- sqrt(b² - 4ac)) ÷ 2a

And get x = (54 +/- 36) ÷ 90

And the x = 1 root is okay, i know that one is right because of the solution and i know its false. But my question is about x = 0.2 since that root doesnt exist in the solution since they did the approach by rearanging like Denis suggests, i also did that first but it hit me that i might get another answer by doing it the long hard way..

And they teach us that you hvae to find all possible roots and try them if they exist. But since this is an old exam im wondering if im doing something wrong, or if the solution is wrong.

When you square both sides you can get extraneous solutions. Your arithmetic seems to be correct to me......

 

[Ishuda]

Sorry but i really dont understand how to write this properly from android.

I have an equation.
3sqrt(x) + sqrt(3 + x) = sqrt(x)
...

We have a problem here I think. First, rearrange the equation
2 sqrt(x) = - sqrt(3 + x)
So if by the sqrt(x) you mean the usual positive square root of a number, this equation has no solutions since the largest the right hand side can be is -sqrt(-3) since x must be non-negative because of the left hand side.