In this section, we want to start our work with square roots. I've written the definition over here on the board for the square root.
If x is a positive number, then this stands for the square root of x; and it is the positive number we square in order to get x.
So, for instance, the √(9) = 3 because 3² = 9; likewise, the square root of √(121) = 11 because 11² = 121.
Now, we have cubed roots over here, which are numbers that we cube to get what's inside there.
But first let's go to the board and look at some square root problems. It says, "Simplify"; and the first one I have is √(81).
Well, that's the number that I square to get 81, that's what this stands for: the positive number I square to get 81.
That's 9 because 9² = 81. So, √(81), or the positive square root of 81, is 9 because 9 to the second power gives us 81.
So, square roots really are the reverse of exponents. Let's look at our next one. Let's simplify 16√(9).
Well, this will be 16 times, this stands for the number that I square to get 9; that number is 3, and then, 16(3) = 48.
So, this is multiplication: 16√(9) stands for multiplication. These aren't connected together;
it's simply multiplication when I don't see the operation symbol shown there. The positive square root of 9 is 3; 16(3) = 48.
Here's our next one: √(49) + √(64). Let's simplify: √(49), this is the positive number I square to get 49, that's 7;
plus √(64), this is the positive number I square to get 64, that's 8. 7 + 8 = 15. So, that simplifies very nicely down to just 15.
So, √(49) + √(64) = 7 + 8; and that comes out to be 15. Let's try another problem similar to that:
√(16) - √(9); this is the positive number I square to get 16; that's 4; minus, that's that subtraction sign,
the positive number I square to get 9, which is 3. 4 - 3 will be the number 1. So, nice simple problem here,
just using the definition for these square roots. Okay, let's go on to problem number 5 here; it's a little bit more complicated:
3√(25) - 9√(49). So, this will be 3 times, square root 25 is 5;
minus 9 times, the positive square root of 49 is 7. 3(5) = 15 minus, 9(7) = 63, which is -48.
Okay, so I simplified each of these roots right here. Then I do my multiplication, and then finally I do my subtraction to end up with 48.
Our next problem is a square root that involves a fraction. Let's find the number we square to get 16/49.
Well, this will be the number that we square to get 16 in the numerator and 49 in the denominator.
How about 4/7? (4/7)(4/7) = 16/49. So, 4/7 is the answer to that √(16/49) = 4/7. Our next problem: √(100/121).
Let's see. The number I would square to get this fraction will be 10 over, let's see, the number I square to get 121 is 11; so, 10/11.
So, I can take square roots of fractions just like I can take square roots of whole numbers.
I simply look for the number that I square to get what's under that square root sign.
Let's look at a couple of problems now where we approximate some square roots. Problem number 8, I have 5√(5).
And I want to approximate this with a decimal. So, I'm going to use the calculator to look up the square root of 5. So, 5 square root: I get 2.236.
Now, that's just to the nearest thousandth. Now, I want to take that and multiply by 5; and that comes out to be 11.180.
So, if I have a calculator, I can simply press that 5, enter 5, and then press the square root button,
and I'll end up with a decimal approximation to that square root. Once I have that, I can use the calculator and multiply that by 5 and get this.
Let's try a decimal approximation for this right here: √(18). I'll do that on the calculator; and let's just go to the nearest thousandth again.
This is 4.243 plus, now, √(19); and that comes out 4.359. So 4.243, 4.359, and then let's add those up: 4.243 + 4.359,
and that gives me 8.602. Now, that's the decimal approximation for the sum of these two things right here.
Now a real common mistake that I see students make all the time when they work with square roots is this.
They'll take √(18) + √(19), and they'll write it as square root, or they'll think of it as, √(18 + 9).
There is no property of square roots that allows us to add square roots this way.
And one way to see that this won't work is to take 18 + 19 and then find its square root. You'll see that you don't get 8.602. Let's try it.
So, I'll take 18 + 19 = 37. When I take the square root, I get 6.08. 6.08 = √(18 + 19).
So, you can see that's not the same as the sum of the square root of 18 plus the square root of 19. Now, one last thing to do in this section,
I want to go over to the other board here and take a look at what's called Pythagorean theorem.
The Pythagorean theorem has to do with right triangles; and what is says is
that the square of the longest side is always equal to the sum of the squares of the other two sides. Another way to say the same thing is
that this longest side, which is called the hypotenuse, is always equal to the square root of the sum of the squares of these other two sides,
right here. So, what we want to do is go to the board and work a problem that involves one of these right triangles.
So, I've written down here, "Find x," and I have a right triangle in which the hypotenuse I've labeled with x. One side is 5, and the other side is 12.
Well, by the Pythagorean theorem, I know that x =√(5² + 12²).
So, I have the √(5² + 12²). That will be, the square root of, 5² = 25, + 12² = 144.
So, I have 25 + 144. If I add those up, see, I get the √(169).
And √(169), we know to be 13. So, if one side is 5, the other side is 12, in a right triangle,
then the hypotenuse must be equal to 13 according to the Pythagorean Theorem.
So, the Pythagorean theorem is a theorem that's been around for two thousand years that involves the square root.