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Q: Let A equals 1 plus 2i and let C equals 5 plus 5i what is A-C?

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x2 +x=3x-5 so x2 -2x+5=0 which does not factor ( over the real numbers) so you can either complete the square of use the quadratic equation to solve. Let's complete the square. (x-1)2 =-4 x-1= plus of minus 2i so x=1+2i or x=1-2i Now check it just as you would a real answer. 1+2i -1 is 2i and 2i squared is -4 as desired Now 1-2i-1 is -2i and (-2i) squared is -4 also. We know the answer is not real if we simply calculated the discriminant. b2 -4ac=4-4(5)<0 so there are no real answers as we found.

y equals x plus 4 when y equals 0 then x equals 2i i is the square root of negative 1

Using the quadratic formula, you get the complex answers of 1 + 2i and 1 - 2i

10 + 6i and 7 + 2i = 10 + 6i + 7 + 2i = 17 + 8i

1.75 + 1.75i7i/2 + 2i = 3.5i + 2i = 5.5i

It is 3 minus 2i

5 + 2 i 2 -3 i = 0(i + 1)(2i - 5) = 0(i + 1) = 0, so, i = -1and(2i - 5) = 0, so, i = 5/2i two values are: -1 and 5/2

The four roots are:1 + 2i, 1 - 2i, 3i and -3i.

x^2 plus 4 = 0x^2 = -4square root both sidesx = the square root of -4x = 2i

(-2 + 3i) + (-1 - 2i) = -2 + 3i - 1 - 2i = -2 - 1 + 3i - 2i = -3 + i

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