How would you simplify these radicals completely ??

Question

a. √50 b. √32 / 4 c. √2 • √10 d.   e. f. -√12 - 2 √3 - 2 √20

Answer 1

When simplifying radicals, remember to "pull out" all perfect squares from under the square root. The trick is to find which factors are perfect squares, because those terms can be pulled out.

a.

√(50)

50 is the same as 2*25

√(50) = √(2*25)

25 is equal to 5², which makes it a perfect square. So take the square root of 25 (which is 5), and pull it out of the radical:

=5√(2)

b.

√(32)/4 = √(16*2)/4

16 is a perfect square:

= 4√(8)/4

now the 4 in the numerator cancels with the 4 in the denominator:

= √(8)

c.

When multiplying square roots, just multiply the terms under the radical, then simplify:

√2 * √10

= √(20)

= √(4*5)

= 2√(5)

d.

This is the same as above, but remember to multiply the numbers in front of the radicals:

6√(2) * 2√(2)

= (6*2) (√2 * √2)

= 12*2

= 24

e.

When adding and subtracting terms with radicals, you can treat the radicals like a variable.

1√3 + 2√3 - 7√3

Here we can factor a √3 out of each term:

= (1+2-7)√3

= (-4)√3

f.

-√(12) - 2√(3) - 2√(20)

This problem uses all of the techniques we've used so far. The first step is to simplify all of the terms:

= -√(4*3) - 2√(3) - 2√(4*5)

= -2√(3) - 2√(3) - 2*2√(5)

= -2√(3) - 2√(3) - 4√(5)

Notice that now we have two terms with a √3. This means we can easily add them together:

= -4√(3) - 4√(5)

Answer 2

Hi Luhan! To answer all of the parts, you will need to know prime factorization.

Note: sqrt(a) is the square root of a!

Prime factorization is a way to break down numbers into prime numbers.

For example, for the number 24, 24 = 2 * 2 * 2 * 3 = (2 ^ 3) * 3 (notice how 2s and 3s are prime numbers).

Another rule to know is that sqrt(a * b) = sqrt(a) * sqrt(b), given that a and b are numbers.

A) Break down 50 into prime numbers. 50 = 2 * 5 * 5 = 2 * (5 ^ 2).

sqrt(50) = sqrt(2 * (5 ^ 2)) = sqrt(2) * sqrt(5^2) = sqrt(2) * 5 <<< ANSWER

B) Break down 32 into prime numbers. 32 = 2 * 2 * 2 * 2 * 2 = (2 ^ 2) * (2 ^ 2) * 2

sqrt(32) = sqrt( (2 ^ 2) * (2 ^ 2) * 2 ) = sqrt(4) * sqrt(4) * sqrt(2) = 2 * 2 * sqrt(2) = 4 * sqrt(2).

Therefore, sqrt(32) / 4 = 4 * sqrt(2) / 4 = sqrt(2) <<< (ANSWER)

C) sqrt(2) * sqrt(10) = sqrt(2) * sqrt(2 * 5) = sqrt(2 * 2) * sqrt(5) = 2 * sqrt(5) <<<<< (ANSWER)

D) 6 * sqrt(2) * 2 * sqrt(2) = 6 * 2 * sqrt(2) * sqrt(2) = 12 * sqrt(4) = 12 * 2 = 24 <<< (ANSWER)

Before I move on to E and F, let me go over a rule for adding/subtracting radicals.

If there are variables or numbers that ⇒ ab + cb + gb (note that b is in every term), then we can group them as ab + cb + gb = (a + c + g) * (b). Using this rule will help you solve E and F.

E) 1 * sqrt(3) + 2 * sqrt(3) - 7 * sqrt(3) = (1 + 2 - 7) * sqrt(3) = (-4) * sqrt(3) <<<<< (ANSWER)

Part F will combine the rules of prime factorization and the group factoring together. Additional steps are added for clarity.

F) -1 * sqrt(12) - 2 * sqrt(3) - 2 * sqrt(20)

Prime factorize sqrt(12) and sqrt(20) (you can't further simplify sqrt(3) because 3 is a prime number).

sqrt(12) = sqrt(2 * 2 * 3) = sqrt(2 * 2) * sqrt(3) = sqrt(4) * sqrt(3) = 2 * sqrt(3)

sqrt(20) = sqrt(2 * 2 * 5) = sqrt(2 * 2) * sqrt(5) = sqrt(4) * sqrt(5) = 2 * sqrt(5)

Using this, rewrite -1 * sqrt(12) - 2 * sqrt(3) - 2 * sqrt(20) = -1 * 2 * sqrt(3) - 2 * sqrt(3) - 2 * 2 * sqrt(5)

= -2 * sqrt(3) - 2 * sqrt(3) - 4 * sqrt(5) = (-2 - 2) * sqrt(3) - 4 * sqrt(5) = -4 * sqrt(3) - 4 * sqrt(5) <<< (ANSWER)

I hope this helped you out! Please let me know if you need additional help! Thanks!