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Algebra

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Fill in the blanks with real numbers:

(sqrt(3) + i)*6*(6 + i)^4 = ___ + ___ i.

Dec 19, 2023

#1
+259
+1

Here's how to fill in the blanks:

Expand the second term using the Binomial Theorem:

(6 + i)^4 = 6^4 + 4 * 6^3 * i + 6^2 * i^2 + 4 * 6 * i^3 + i^4

= 1296 + 1440i - 36 + 144i - 1

= 1260 + 1584i

Multiply the first term by the expanded second term:

(sqrt(3) + i) * (1260 + 1584i) = 1260 * (sqrt(3) + i) + 1584i * (sqrt(3) + i)

Calculate each product separately:

1260 * (sqrt(3) + i) = 1260 * sqrt(3) + 1260i

1584i * (sqrt(3) + i) = 1584i * sqrt(3) + 1584i^2

Simplify the imaginary unit multiples:

1584i^2 = -1584

Substitute the simplified terms back into the expression:

1260 * sqrt(3) + 1260i + 1584i * sqrt(3) - 1584

Combine like terms:

(1260 + 1584) * sqrt(3) + (1260 + 1584)i

Fill in the blanks:

Therefore, the expression equals (6194.08) + (15215.54)i

Dec 19, 2023
#2
+126978
0

(sqrt 3 + i)*6  =  6sqrt 3 +  6i  = a

(6 + i)^4  =

6^4 + 4*6^3 * i  + 6*6^2* i^2  + 4*6*i^3 + i^4  =

1296 + 864i - 216 -24i + 1 =

1081 + 840i  =  b

(a)(b)  = (6sqrt3 + 6i) ( 1081 + 840i)  =  6486sqrt 3 + 6486 i + 5040sqrt(3) i - 5040 =

6486sqrt (3) -5040 + (6486 i + 5040sqrt (3) ) i

Dec 19, 2023