Solving square-root equations: two solutions | Mathematics III | High School Math | Khan Academy

Let's say that we have the equation ( 6 + 3w = \sqrt{2w + 12} + 2w ).

See if you can pause the video and solve for ( w ), and it might have more than one solution, so keep that in mind.

All right, now let's work through this together. The first thing I'd like to do whenever I see one of these radical equations is just isolate the radical on one side of the equation. So let's subtract ( 2w ) from both sides. I want to get rid of that ( 2w ) from the right-hand side. I just want the radical sign. If I subtract ( 2w ) from both sides, what am I left with? Well, on the left-hand side, I am left with ( 6 + 3w - 2w ). Well, ( 3 ) of something take away ( 2 ) of them, you're going to be left with ( w ).

So, ( 6 + w = \sqrt{2w + 12} ).

Now, to get rid of the radical, we're going to square both sides. We've seen before that this process right over here is a little bit tricky because when you're squaring a radical in a radical equation like this and then you solve, you might find an extraneous solution. What do I mean by that? Well, we're going to get the same result whether we square this or whether we square that because when you square a negative, it becomes a positive. But those are fundamentally two different equations.

We only want the solutions that satisfy the one that doesn't have the negative there. So that's why we're going to test our solutions to make sure that they're valid for our original equation.

If we square both sides, on the left-hand side we're going to have ( (6 + w)^2 ). It's going to be ( w^2 + 2(6)(w) + 6^2 ). So, ( w^2 + 12w + 36 ) is equal to ( 2w + 12 ).

Now we can subtract ( 2w ) and ( 12 ) from both sides. So let's do that, so then we can get it into kind of a standard quadratic form.

So let's subtract ( 2w ) from both sides and let's subtract ( 12 ) from both sides. Once again, I just want to get rid of this on the right-hand side. I am going to be left with ( w^2 + (12w - 2w) + (36 - 12) = 0 ).

So, ( w^2 + 10w + 24 = 0 ). Let's see, to solve this, is this factorable? Are there two numbers that add up to ( 10 ) and whose product is ( 24 )? Well, what jumps out at me is ( 6 ) and ( 4 ).

So we can rewrite this as ( (w + 4)(w + 6) = 0 ).

If I have the product of two things equaling zero, to solve this, either one or both of them could be equal to zero. Zero times anything is going to be zero. So, ( w + 4 = 0 ) or ( w + 6 = 0 ).

Over here, if you subtract ( 4 ) from both sides, you get ( w = -4 ) or subtract ( 6 ) from both sides here, ( w = -6 ).

Now, let's verify that these actually are solutions to our original equation. Remember, our original equation was ( 6 + 3w = \sqrt{2w + 12} + 2w ).

So let's see if ( w = -4 ) works.

If ( w = -4 ), that gives us ( 6 + 3(-4) = \sqrt{2(-4) + 12} + 2(-4) ).

So this would be ( 6 - 12 = \sqrt{-8 + 12} - 8 ).

This simplifies to ( -6 = \sqrt{4} - 8 ), or ( -6 = 2 - 8 ).

That indeed holds true, ( -6 = -6 ).

So this is definitely a solution.

Now, let's try ( w = -6 ).

So if ( w = -6 ), we get ( 6 + 3(-6) = \sqrt{2(-6) + 12} + 2(-6) ).

This gives us ( 6 - 18 = \sqrt{-12 + 12} - 12 ).

So we have ( -12 = 0 - 12 ), which is also true.

So we get ( -12 = -12 ).

Therefore, these are actually both solutions to our original radical equation.