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- Let's do one more radical equation example.
- And we'll make this one a slightly hairy one.
- So let's say we have the principal square root of 10
- minus 5x plus the principal square root of 1 minus x is
- equal to 7.
- Now, just like I've said over the last two videos, your goal
- is always to isolate one of the radicals and then square
- both sides of the equation.
- So we can isolate this one right here by subtracting this
- radical from both sides of this equation.
- So you get minus the square root of 1 minus x, minus the
- square root of 1 minus x.
- The left-hand side, these cancel out.
- The left-hand side, you're just left with the square root
- of 10 minus 5x is equal to-- on the right-hand side, you
- have 7 minus the square root of 1 minus x.
- All I did is subtracted this from both sides, essentially
- moving it over to the right-hand side.
- Now we've isolated one of the radicals.
- Let's square both sides of this equation.
- So this squared, 10 minus 5x squared, is just going to be--
- or the square root of 10 minus 5x squared, so if I square
- that, that's just going to be equal to 10 minus 5x.
- And this is going to be equal to this expression squared.
- Now, this might look a little daunting, but you just have to
- remember that a plus b squared is equal to a squared plus 2ab
- plus b squared.
- In fact, you don't have to remember it.
- You could actually multiply it out if you ever forget this.
- Now, this is just the same situation where a is 7, and
- our b is negative the square root of 1 minus x.
- So if we expand this out or multiply this out, this is
- going to be 7 squared, which is 49.
- And then plus 2 times 7 times this.
- So this is a negative right here, so it's going to be
- minus 2 times 7 times the square root of 1 minus x.
- So we've got the negative side from that negative sign, the 7
- is over there, and 1 minus the square root of 1 minus x is
- there, 2 times it.
- When you take a perfect square, that just ends up,
- when you actually take the product.
- And then finally, plus this guy squared.
- This guy over here squared.
- Now the negative part, negative 1 squared, that just
- becomes positive.
- Square root of 1 minus x squared is just going
- to be 1 minus x.
- And let's simplify this.
- So this is going to be equal to 10 minus 5x is equal to--
- well, one thing we can do, let's add the 49 and the 1.
- So you add the 49 and the 1, you get 50, and then minus--
- this is 2 times 7 is 14 times the square root of 1 minus x.
- And then we have this final minus x out here, minus x.
- And let's see what else we can do here.
- We could subtract 10 from both sides of this equation.
- I don't want to do too many steps at once, because this is
- a slightly involved problem.
- So if you subtract 10 from both sides, you get negative
- 5x is equal to 40 minus 14 times the square root of 1
- minus x minus x.
- And then we can add x to both sides, essentially getting rid
- of that character over there.
- So if you add x to both sides of this equation, the
- left-hand side becomes negative 4x, and the
- right-hand side, this guy disappears, and you're just
- left with 40 minus 14 times the square root of 1 minus x.
- And now let's subtract 40 from both sides.
- I'm just trying to isolate this expression right here.
- So let's subtract 40 from both sides.
- We get negative 4x minus 40 is equal to negative 14 times the
- square root of 1 minus x.
- Just subtracted this from both sides, so it disappeared on
- the right, shows up in negative form on the left.
- Now, we could square both sides, but let's simplify it a
- little bit.
- Let's divide both sides by negative 2.
- That'll get rid of all of these negative numbers and
- make the numbers a little bit smaller.
- So you divide everything by negative 2.
- We get 2x-- let me switch colors.
- We get 2x plus the 20 is equal to 7 times the square
- root of 1 minus x.
- Just divide everything by negative 2.
- 40 divided by negative 2 is 20, 14 divided by negative 2
- is positive 7.
- Now we could divide by 7, but that makes things awkward.
- We've essentially isolated the radical.
- We're scaling it by a little bit.
- But if we square both sides of this equation, that radical
- will disappear.
- So let's square both sides of that equation.
- The left-hand side becomes 4x squared plus 2x time 20 is 40x
- times 2 is 80x plus 400 is going to be equal to 7
- squared, which is 49, times this expression squared.
- Well, the square root of 1 minus x squared is just going
- to be 1 minus x.
- Let's just keep going down here.
- I have space down here.
- I can just keep going.
- So what can we do here?
- I don't want to skip too many steps.
- So 4x squared plus 80x plus 400 is equal to--
- distribute the 49.
- 49 minus 49x.
- And now, let's see, we could add 49x to both
- sides of this equation.
- Actually, let's do this.
- Let's add 49x to both sides, and let's subtract 49 from
- both sides of the equation.
- And we are left with-- the left-hand
- side becomes 4x squared.
- Now, 80 plus 49 is-- 80 plus 40 is 120, so it's going to be
- 129, plus 129x.
- And then 400 minus 49 is 351, plus 351 is equal to-- these
- guys cancel out, those guys cancel out-- is equal to 0.
- Now, let me just make sure that I haven't made some crazy
- mistake because these are strange-looking numbers.
- 80 plus 49-- yep, that's right.
- 129.
- 400 minus 49, that's 351.
- Now we have just a straight-up quadratic equation.
- Let's use the quadratic formula since we have all
- these crazy numbers.
- So our solutions, we're going to get x is
- equal to negative b.
- So it's negative 129 plus or minus the square root of 129
- squared minus 4 times a, which is 4, times c, which is 351--
- definitely going to need a calculator for this one-- all
- of that over 2 times a, which is 2 times 4, which is 8.
- Well, let's get our calculator out to
- actually calculate this.
- So let's figure out the discriminant first. So we have
- 129 squared minus-- we could say 16 times 351.
- Nope, that's not what I wanted to do.
- 16 times 351.
- 4 times 4 is 16, times 351, is that crazy number.
- Let's take the square root of it.
- And on this calculator, you press second and
- answer right here.
- That'll use the last answer, so it's going to take the
- square root of that right there.
- It's 105.
- So lucky for us, it's not that crazy of a number.
- So we get x is equal to negative 129 plus or
- minus 105 over 8.
- Let's just use a calculator for this.
- So the first one, if we have a negative 129 plus
- 105, what do we get?
- Negative 24, then you divide that by 8, that's going to be
- negative 3.
- So one of our answers is negative 3.
- Now, what happens if we take negative 129 minus 105?
- We get negative 234.
- Divide that by 8.
- We get negative 29.25.
- So our two answers are negative 3 and negative 29.25.
- So x is equal to negative 3 and x is
- equal to negative 29.25.
- Now, remember, this is a radical equation.
- We have to validate that both of these work, or if they
- don't work, they're going to be extraneous solutions, or
- one of them might be an extraneous solution.
- So let's copy these and go back up to our original
- problem up here, right there.
- Let me paste them, just so we remember what our solutions
- were, and let's just evaluate them in the calculator, to see
- if they work.
- I don't want to use that calculator.
- I'll Use this calculator.
- Much better.
- All right.
- So let's try negative 3.
- So we're going to take the square root of 10 minus 5
- times negative 3-- that's that right there; we got out of the
- radical-- plus the square root of 1 minus negative 3.
- I'm doing nothing in my head.
- Obviously, that would be positive 3, and all of that.
- What is that going to be equal to?
- It's equal to 7.
- So negative 3 definitely works.
- Now let's try negative 29.25 and see what happens.
- So we take-- once again-- the square root of 10 minus 5
- times negative 29.25 right?
- Close parentheses plus the square root of 1 minus
- negative 29.25 is equal to 18.
- This didn't work.
- It's not even equal to negative 7.
- That's because we took the principal square roots twice
- or we squared it twice, or we lost information twice.
- But the simple answer is this does not work.
- It is an extraneous solution.
- We got that solution assuming not the principal roots,
- assuming the negative square roots somewhere along here.
- So for this equation that deals with just the principal
- roots, we can say that this right here is
- an extraneous solution.
- And the only valid solution to this equation is x is equal to
- negative 3.

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