# Complex numbers and square root property

• April 28th 2008, 03:12 PM
rowdy3
Complex numbers and square root property
Im doing test corrections and need some help. My answers are wrong. Could you please show your work? Thanks.

simpify

square root of -12
I took the i out. I see that 3 and 4 go into 12. I reduced the 4 to 2. I got 3i square root of 2

square root of -16 and the square root of -49
Took the i out and got 4. Took the i out and got 7. My answer was 2i square root of 2 i square root of 7.

Divide the complex numbers
4 - 3i / 2 - 7i
I added 2 + 7i on the top. I foiled the problem. and got 8+28i - 6i - 21i^2. My answer was 29 - 22i

Solve the following equations by using the square root property.

4x^2 + 16 = 0

Thanks.
• April 28th 2008, 03:16 PM
Mathstud28
Quote:

Originally Posted by rowdy3
Im doing test corrections and need some help. My answers are wrong. Could you please show your work? Thanks.

simpify

square root of -12
I took the i out. I see that 3 and 4 go into 12. I reduced the 4 to 2. I got 3i square root of 2

square root of -16 and the square root of -49
Took the i out and got 4. Took the i out and got 7. My answer was 2i square root of 2 i square root of 7.

Divide the complex numbers
4 - 3i / 2 - 7i
I added 2 + 7i on the top. I foiled the problem. and got 8+28i - 6i - 21i^2. My answer was 29 - 22i

Solve the following equations by using the square root property.

4x^2 + 16 = 0

Thanks.

if a>0
$\sqrt{-a}=i\sqrt{a}$
• April 28th 2008, 03:26 PM
rowdy3

√(-12)
= √(12i^2)
= i√12
= i(√3 x √4)
= i(√3 x 2)
= 2i √3

√(-16)
= √(16i^2)
= i√16
= 4i

√(-49)
= i√49
= 7i

(4 - 3i) / (2 - 7i)
= (4 - 3i)(2 + 7i) / (2 - 7i)(2 + 7i)
= (8 + 28i - 6i - 21i^2) / (4 - 49i^2)
= (8 + 22i + 21) / (4 + 49)
= (29 + 22i) / 53
= 29/53 + (22/53)i

4x^2 + 16 = 0
4(x^2 + 4) = 0
x^2 + 4 = 0
x^2 = -4
x = √-4
x = √(4i^2)
x = 2i
• April 28th 2008, 03:32 PM
Mathstud28
Quote:

Originally Posted by rowdy3

√(-12)
= √(12i^2)
= i√12
= i(√3 x √4)
= i(√3 x 2)
= 2i √3

√(-16)
= √(16i^2)
= i√16
= 4i

√(-49)
= i√49
= 7i

(4 - 3i) / (2 - 7i)
= (4 - 3i)(2 + 7i) / (2 - 7i)(2 + 7i)
= (8 + 28i - 6i - 21i^2) / (4 - 49i^2)
= (8 + 22i + 21) / (4 + 49)
= (29 + 22i) / 53
= 29/53 + (22/53)i

4x^2 + 16 = 0
4(x^2 + 4) = 0
x^2 + 4 = 0
x^2 = -4
x = √-4
x = √(4i^2)
x = 2i

Good job! (Clapping)
• April 28th 2008, 04:57 PM
rowdy3
Which of the following is not equal to i
a. i^9
b. i^21
c i ^33
d. i^51

I picked B.
• April 28th 2008, 05:10 PM
o_O
Knowing that:
$i^{1} = i$
$i^{2} = - 1$
$i^{3} = i^{2}i = -i$
$i^{4} = i^{2}i^{2} = 1$
$i^{5} = i^{4}i = i$

The pattern repeats for powers 1 - 4. So:
If the exponent is 21, then 21/4 = 5 remainder 1 (i.e. go through 5 cycles plus 1 increment). This gives us: $i^{21} = i^{1} = i$

So it isn't B but perhaps you can use the same method to determine which of the answers do not equal to i.
• April 28th 2008, 05:52 PM
finch41
Quote:

Originally Posted by rowdy3

√(-12)
= √(12i^2)
= i√12
= i(√3 x √4)
= i(√3 x 2)
= 2i √3

√(-16)
= √(16i^2)
= i√16
= 4i

√(-49)
= i√49
= 7i

(4 - 3i) / (2 - 7i)
= (4 - 3i)(2 + 7i) / (2 - 7i)(2 + 7i)
= (8 + 28i - 6i - 21i^2) / (4 - 49i^2)
= (8 + 22i + 21) / (4 + 49)
= (29 + 22i) / 53
= 29/53 + (22/53)i

4x^2 + 16 = 0
4(x^2 + 4) = 0
x^2 + 4 = 0
x^2 = -4
x = √-4
x = √(4i^2)
x = 2i

should not the last one be plus or minus 2i?
• April 28th 2008, 06:41 PM
rowdy3
It would be c.

Solve the equation by completing the quare and applying the square root property.

w^2 - 16w = 3
Take half of b and square it. b = -16. -16/2= 8. 8^2=64.
w^2- 16w + 64= 67
Factor the Left side
(w-8)^2= 67
Using the Square root property take the square root of both sides.
w-8 or square root of 67
w= 8 or the square root of 67
• April 29th 2008, 03:17 PM
rowdy3
Complex numbers
(2+4i) - (6 - 2i) + (3 - 5i)
I got - 4 -10i

Solve the equation by using the square root property

x^2 - 30 = 0
Do you divide 30 by 2 and times the 2?

(y-6)^2 + 4 = 0

Thanks
• April 29th 2008, 03:29 PM
rowdy3
(2+4i) - (6 - 2i) + (3 - 5i)
2-6+3 = -1
4+2-5 = 1
-1 + i

x^2 - 30 = 0
x^2 = 30
x = +/-sqrt(30)

(y-6)^2 + 4 = 0
(y-6)^2 = -4
y-6 = +/- 2i
y = 6 + 2i or 6 - 2i