More Involved Radical Equation Example

⇐ Use this menu to view and help create subtitles for this video in many different languages. You'll probably want to hide YouTube's captions if using these subtitles.

More Involved Radical Equation Example

相關課程

0 / 750

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.

留言:

載入中...

新增一則留言(尚未登入)

more involved radical equation example 討論區:

載入中...