Solving an equation, I have an incorrect solution

[Alevelman]

I have been attempting to solve the equation, I have arrived at two roots however only one of them seems to be correct. If possible could you spot a possible mistake I have made.

$\sqrt{3x + 10} =2 + \sqrt{x+4}$

$3x + 10=4+4\sqrt{x+4}+x+4$

$0.5x+0.5=\sqrt{x+4}$

$0.25x^2+0.5x+0.25=x+4$

$x^2+2x+1=4x+16$

$x^2-2x-15=0$

$(x-5)(x+3)=0$

$x=5$ and $x=-3$

However when you put the results back into the equation only x=5 is true
Thanks for any help

 

[Stephen Tashi]

In your first step, you squared both sides of an equation. When this is done, you may create an equation that has more solutions than the original equation. ( For example, consider the equation y = 5. If we square both sides to obtain y^2 = 25 the second equation has solutions y = 5 and y = -5.)

If you have a very sophisticated mathematical mind and some verbal skills, you can write notes in your work indicating the possible places where extra solutions may be introduced and how to intend to eliminate them. Most people don't use that approach. They simply crank out the problem and check whether the final solutions that are produced actually work in the original equation. That is what you did. The fact that not all solutions work doesn't mean you made any error in the algebraic manipulations.

 

[Marcus27]

I have been attempting to solve the equation, I have arrived at two roots however only one of them seems to be correct. If possible could you spot a possible mistake I have made.

$\sqrt{3x + 10} =2 + \sqrt{x+4}$

$3x + 10=4+4\sqrt{x+4}+x+4$

$0.5x+0.5=\sqrt{x+4}$

$0.25x^2+0.5x+0.25=x+4$

$x^2+2x+1=4x+16$

$x^2-2x-15=0$

$(x-5)(x+3)=0$

$x=5$ and $x=-3$

However when you put the results back into the equation only x=5 is true
Thanks for any help

There are sometimes roots which do not satisfy the original equation these are called 'extraneous solutions'. Khan academy has videos explaining this far better than I ever could.

Edit: I read equation wrong :(

 

[Ray Vickson]

It seems in step 1 you squared (2 + (x + 4)^1/2)) rather than squaring each term separately.

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I hope you realize that he had to square the whole expression $2 + \sqrt{x+4}$, rather than squaring each term separately. Doing the latter would have been wrong!

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There are sometimes roots which do not satisfy the original equation these are called 'extraneous solutions'. Khan academy has videos explaining this far better than I ever could.

 

[Marcus27]

Yeah I realized after I posted, I edited my post so I did not to confuse anybody, I still have lots of mathematics to learn myself but I thought I could help this guy but made a mistake myself

 

[SammyS]

I have been attempting to solve the equation, I have arrived at two roots however only one of them seems to be correct. If possible could you spot a possible mistake I have made.

$\sqrt{3x + 10} =2 + \sqrt{x+4}$
...

$0.25x^2+0.5x+0.25=x+4$
...

$x=5$ and $x=-3$

However when you put the results back into the equation only x=5 is true
Thanks for any help

As Marcus points out, x = -3 is an "extraneous solution".

It is a solution to the equation $$ \sqrt{3x + 10} =2 - \sqrt{x+4}\ \ .$$

If you follow the same set of steps with this equation that you did with yours, you will eventually arrive the equation: $$0.25x^2+0.5x+0.25=x+4\ \ $$
just as you did with your equation.