Find the product of the complex number: (5-2i)(5+2i)?

Question

Also:

Solve the equation for solution and check for extraneous solutions:

√2y^2-1=y

Answer 1

The product of two conjugate complex numbers is real:

(5 - 2i) (5 + 2i) = 5 * 5 + 5 * 2i - 5 * 2i - 2i * 2i
... = 25 + 10i - 10i - 4i^2
... = 25 + 4 = 29

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sqrt(2y^2 - 1) = y -------- take squares (introduces extraneous sol.)

2y^2 - 1 = y -------- rearrange

2y^2 - y - 1 = 0 -------- solve using quadratic formula

y = 1/4 +/- 1/4 sqrt 9
y = 1 or y = -1/2

Check:
sqrt (2 * 1^2 - 1) = sqrt 1 = 1, so y = 1 is solution
sqrt (2*(-1/2)^2 - 1) = sqrt (-1/2) <> -1/2 --> extraneous solution

Answer 2

Set up the problem just like you would long multiplication, like you learned in the 3rd grade:

.... 5 - 2i
(x) 5 + 2i
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Now, for each term on the bottom, find the products with the terms on the top. As you list the terms below, group like terms together.

.... 5 - 2i
(x) 5 + 2i
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... 10i+ 4
25-10i
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25- 0i+ 4 =====> 29 + 0i ===> 29

The technique above for setting up a multiplication vertically also works for polynomials.

Also, you may have recognized the problem as being of the form (a + b)(a - b) = a^2 - b^2. Here, a = 5 and b = 2i. Then, a^2 = 25 and b^2 = (2i)^2 = (4)(-1) = -4. So, the answer is 25 - (-4) = 29.

Answer 3

Answer to 1st one is 5^2 + 2^2 = 29.

Answer to second question is:

(1 + sqrt(1 + 4* sqrt(2))) / (2 * sqrt(2))

and

(1 - sqrt(1 + 4* sqrt(2))) / (2 * sqrt(2))

Here sqrt means square root.

Answer 4

29 is absolutely correct