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andriy [413]
3 years ago
5

Which pair of complex factors results in a real-number product?

Mathematics
2 answers:
Crank3 years ago
8 0

we will proceed to solve each case to determine the solution of the problem

we know that

i^{2} =-1

<u>case A</u>. 15*(-15i)

15*(-15i)=-225i

<u>The result of case A is not a real number product</u>

<u>case B</u>. 3i*(1-3i)

3i*(1-3i)=3i*(1)-(3i)*(3i)\\=3i-9i^{2}\\=3i-9*(-1)\\=3i+9

<u>The result of case B is not a real number product</u>

<u>case C</u>. (8+20i)*(-8-20i)

(8+20i)*(-8-20i)=8*(-8)+8*(-20i)+20i*(-8)-20i*(20i)\\=-64-160i-160i-400i^{2}\\=-64-320i-400*(-1)\\=-64-320i+400\\=336-320i

<u>The result of case C is not a real number product</u>

<u>case D</u>. (4+7i)*(4-7i)

(4+7i)*(4-7i)=4^{2} -(7i)^{2}\\=16-49i^{2}\\=16-49*(-1)\\=65\\

<u>The result of case D is a real number product</u>

therefore

<u>the answer is the option</u>

(4+7i)*(4-7i)



defon3 years ago
4 0
To answer the problem above, we must first define what "i" means. i means imaginary number which means square root of - 1, to get rid of i, you simply need to square, in this problem, we need to look for an equation where "i" is squared. The answer is D. (4+7i)(4-7i) to justify:
(4+7i)(4-7i)
(16-28i+28i-49i^2)
(16-49i^2)
16-49 = -33
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