3899 Shares

Solve for x and y in equation with imaginary numbers Topic: Problem solving with answer and solutions
June 17, 2019 / By Bennett
Question: solve for x and y where x and y are real numbers. i is imaginary There are no examples in the book. The answer in the back of the book for the following question is x= 4 y= -16 I need to know HOW to solve it 8 (3x y)i =2x-4i for some reason yahoo does not display plus sign. equation is 8 plus (3x plus y)i=2x-4i christian, the x=4 and y=-16 were not given in the problem. Those are the answers in the back of the book. And "i" is not a variable, its an imanginary number. i = the square root of -1 gravitative, I dont quite understand why that works. Nothing in my book separates the imaginary numbers in an equation to be separate from the real numbers into 2 different equations. Im still confused. Best Answers: Solve for x and y in equation with imaginary numbers Abiah | 10 days ago
Yahoo isn't displaying plus signs on initial posts due to troubles with a recent update, but it will display them in "additional details" or answers. ---- 8 + (3·x + y)·i = 2·x - 4·i The real parts and the imaginary parts have to be equal. The only real part of the left side is 8, and on the right you have 2x: 8 = 2·x x = 4 Use this value for the imaginary parts, which are also equal: (3·4 + y)·i = -4·i (12 + y)·i = -4·i 12 + y = -4 y = -16 ---- The trick is that it says "x and y are real numbers". If you multiply, add, subtract, or divide real numbers with real numbers.. you will get real numbers. If you multiply the imaginary part with a real part, you will still have an imaginary part, so (3x + y)i will be imaginary as long as x and y are real. This means that the only real part on the left is 8. The only real part on the right is 2x. Thus, they have to be equal to each other. The rest will always be imaginary as x and y are real numbers, so they have to be equal to each other. The imaginary unit works differently than a variable, so this problem is kind of like trying to get: 8 - 4i = 8 - 4i This doesn't work with normal variables where i is, but i isn't a variable. It doesn't have differing values, and it add with real parts. It keeps the factors multiplied by it as separate from the rest of the equation. So if you have real parts of an equation and imaginary parts, they would have to equal each other. If x and y didn't have to be real numbers, that method wouldn't work and multiple solutions could exist.
👍 288 | 👎 10
Did you like the answer? Solve for x and y in equation with imaginary numbers Share with your friends

We found more questions related to the topic: Problem solving with answer and solutions Originally Answered: How can I solve the fourth root of -1 with imaginary numbers?
CORRECTION: oops, because it's not the fourth root of i, the problem amounts to: (a + bi)^2 = i which is really the square root of i. dang, i just did the fourth root of i. but this problem could be doen the same way... a^2 + 2abi - b^2 = 0 + 1i (a^2 - b^2) + (2ab)i = 0 + 1i so we would solve the system: a^2 - b^2 = 0 2ab = 1 +++++++++++++++++++++++++++++++++++++++... this is what I had done before i made the correction: (fourth root of i) one way is to use DeMoivre's formula: this will give you an approximate answer. a second way is to use a system: this will give you the exact answer. the fourth root of i is of course imaginary and so must be of the form a + bi thus, (a + b)^4 = 0 + 1i expanded is: a^4 + (4a^3b)i - 6a^2b^2 - (4ab^3)i + b^4 = 0 + 1i grouping the real and imaginary parts we have: (a^4 - 6a^2b^2 + b^4) + (4a^3b - 4ab^3)i = 0 + 1i generating the sytem: a^4 - 6a^2b^2 + b^4 = 0 4a^3b - 4ab^3 = 1 equation 1 be can be written as: (a^2 - b^2)^2 - 4a^2b^2 = 0 which would be (with the principle root): a^2 - b^2 = 2ab the second equation of the sytems can be written as: a^2 - b^2 = 1/(4ab) so we have: 2ab = 1/(4ab) a^2 = 1/(8b^2) a = 1/[2b√2] substitute this into a^2 - b^2 = 2ab to get: 1/(8b^2) - b^2 = 1/√2 (1/8) - b^4 = b^2/√2 rearrange for completing the square on the quadratic of b^2... b^4 + (1/√2)b^2 = 1/8 (b^2 + 1/(2√2))^2 = 1/8 + 1/8 (b^2 + 1/(2√2))^2 = 1/4 b^2 = - 1/(2√2) + 1/2 b^2 = (√2 - 1) / (2√2) b^2 = (2 - √2) / 2 b = √((2 - √2) / 2) using this value of b, we find: a = √((2 + √2) / 2) the exact value of the cube root of i is: √((2 + √2) / 2) + √((2 - √2) / 2) * i Originally Answered: How can I solve the fourth root of -1 with imaginary numbers?
z⁴ = - 1 z = (a + bi) (a + bi)⁴ = - 1 a⁴ - 6 a² b² + b⁴ + i(4 a³ b - 4 a b³) = - 1 system 4 a³ b - 4 a b³ = 0 a⁴ - 6 a² b² + b⁴ = - 1 4ab(a² - b²) = 0 a² - b² = 0 b² = a² → b = ± a a⁴ - 6 a⁴ + a⁴ = - 1 4a⁴ = 1 → a = ± √(1/2) a = -√(1/2), b = - √(1/2) a = -√(1/2), b = √(1/2) a = √(1/2), b = √(1/2) a = √(1/2), b = -√(1/2) z₁ = -1/√2 -i/√2 z₂ = -1/√2 + i/√2 z₃ = 1/√2 + i/√2 z₄ = -1/√2 - i/√2 Trigonometric form is more healthy :) but I didn't know if you know it 4th roots in complex numbers are 4 Originally Answered: How can I solve the fourth root of -1 with imaginary numbers?
a million. (a million-i)^3 which could be rewritten as (a million-i)(a million-i)(a million-i); multiply the trouble-free 2 gadgets of parentheses together, simplify and multiply the completed ingredient by applying applying applying the suited set of parentheses: (a million-i)(a million-i)(a million-i) (a million - i - i + i^2)(a million-i) (a million - 2i - a million)(a million-i) (a million - 2i - a million - i + 2i^2 + i) (-2i + 2i^2) = -2i - 2 2. (-2i)^4 what's -2 to the fourth? sixteen. what's i^4? it incredibly is (-a million)^2, or a million, so the completed ingredient is sixteen*a million = sixteen. Shulammite
first step: you put the number that goes to x where there is x and the number that goes to y where there is y. so 8(3*4*-16)i=2*4-4i second step:you do the multiplications in side of the sing of aggregations so: 8(12-16)i=2*4-4i then: 8(12-16)i=8-4i then: 8(-4)i=8-4i then: 32i=8-4i then 32+8=-i-4i so 40=-5i so 40/5=5i/5 so 8=i
👍 120 | 👎 2 Originally Answered: Help understanding imaginary numbers? Originally Answered: Help understanding imaginary numbers?