Solving Irrational Equations: How to Discard Solutions

[Alex126]

I figured it out >_>

I got a problem with discarding the second solution of this irrational equation:
$-\sqrt {x^2 - 1} + \sqrt {x^2 +3x} = 2$

First I find the domain, which will end up being $x\leq-3$ v $x\geq+1$ since that's the common union of the domains of each square root.

Then I move the first term to the right:
$\sqrt {x^2 +3x} = 2 + \sqrt {x^2 - 1}$

Now I know left and right are positive, so I can elevate both terms to the power of ². I then get:

$x^2 +3x = 4 + x^2 - 1 +4\sqrt {x^2 - 1}$

Do some sums:
$3x - 3 = 4\sqrt {x^2 - 1}$

Now, here's the part I'm unsure about. How do I proceed now? The method I knew was that I can elevate both terms to the power of ² again, but since I don't know the sign of what's on the left then I'll have to verify the solutions (by substituting them in the equation before elevating to ²) and see if the result is right or not.

However, I read that you could do something else too, and that seemed to make sense. In particular, they said that if you know the sign of either one, then you can "impose" a new condition on the other term of the equation and then verify the result from there. In this case, we know that the term on the right must be positive (or null), so the condition would require me to impose that what's on the left must be positive (or null) too. Therefore:

$3x - 3 \geq 0 => x \geq 1$

Then I can elevate both terms to the power of ², and solve the equation $9x^2+9-18x=4x^2-4$
From there I get two results:
x = 1
x = 13/5 (= 2.6)

Both of these look acceptable results, since they are both contained in the domain AND in the condition $x \geq 1$. However, if I substitute the second result (13/5) in the equation $3x - 3 = 4\sqrt {x^2 - 1}$ I don't get an identity, so this second result must not be acceptable.

I don't understand where is the mistake.

Edit: I had forgotten 4²...

If someone still read it, can I get a confirmation that both methods (substituting the results to verify the identity, and imposing a second condition) are always valid?

 

[Mark44]

I figured it out >_>

I got a problem with discarding the second solution of this irrational equation:
$-\sqrt {x^2 - 1} + \sqrt {x^2 +3x} = 2$

First I find the domain, which will end up being $x\leq-3$ v $x\geq+1$ since that's the common union of the domains of each square root.

Then I move the first term to the right:
$\sqrt {x^2 +3x} = 2 + \sqrt {x^2 - 1}$

Now I know left and right are positive, so I can elevate both terms to the power of ². I then get:

$x^2 +3x = 4 + x^2 - 1 +4\sqrt {x^2 - 1}$

Do some sums:
$3x - 3 = 4\sqrt {x^2 - 1}$

Now, here's the part I'm unsure about. How do I proceed now? The method I knew was that I can elevate both terms to the power of ² again, but since I don't know the sign of what's on the left then I'll have to verify the solutions (by substituting them in the equation before elevating to ²) and see if the result is right or not.

Yes, this is the right approach -- square both sides.

However, I read that you could do something else too, and that seemed to make sense. In particular, they said that if you know the sign of either one, then you can "impose" a new condition on the other term of the equation and then verify the result from there. In this case, we know that the term on the right must be positive (or null)

Or zero, not null..

, so the condition would require me to impose that what's on the left must be positive (or null) too. Therefore:

$3x - 3 \geq 0 => x \geq 1$

Then I can elevate both terms to the power of ², and solve the equation $9x^2+9-18x=4x^2-4$
From there I get two results:
x = 1
x = 13/5 (= 2.6)

You have an error in the second value. I get x = -25/7.

Both of these look acceptable results, since they are both contained in the domain AND in the condition $x \geq 1$. However, if I substitute the second result (13/5) in the equation $3x - 3 = 4\sqrt {x^2 - 1}$ I don't get an identity, so this second result must not be acceptable.

I don't understand where is the mistake.

Edit: I had forgotten 4²...

If someone still read it, can I get a confirmation that both methods (substituting the results to verify the identity, and imposing a second condition) are always valid?

Other than your calculation error, I don't see any problems with your second method. I always use the first method you showed, and then verify that the solutions I get satisfy the original equation.

 

[Alex126]

Yea, -25/7. Damn it lol

Thanks.