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The roots of a quadratic equation are 3 2i and 3-2i what is the quadratic equation?
Wiki User
∙ 15y ago
The two roots given are x = 3+2i and x = 3-2i. Therefore x - (3+2i) = 0, and x - (3-2i) = 0. This also implies that [x - (3+2i)]*[x - (3-2i)] = 0. Rewriting: (x - 3 - 2i)(x - 3 + 2i) = 0 Multiplying out: x^2 - 3x + 2ix - 3x + 9 - 6i - 2ix + 6i - 4i^2 = 0 We note that the 2ix's cancel, as well as the 6i's: x^2 - 3x - 3x + 9 - 4i^2 = 0 And finally, noting that i^2 = -1, we combine like terms and get: x^2 - 6x + 13 = 0, which is the required quadratic equation.
Wiki User
∙ 15y ago
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What is the quartic polynomial function with rational coefficients that has roots you and 2i?
(x - u)*(x - u)*(x + 2i)*(x - 2i) = (x2 - 2xu + u2)*(x2 + 4) = x4 - 2x3u + x2(u2 + 4) - 8xu + 4u2
Factor 2x squared plus 6x - 8 equals 72?
2x2 + 6x - 8 = 72 ∴ 2x2 + 6x + 64 =0 ∴ x2 + 3x + 32 = 0 This can not be factored, as x is not equal to any integer. Using the quadratic equation, we find that: x = -3/2 ± √119 / 2i
What is the conjugate of 7 - 2i?
7 + 2i
What multiplies to make 3 and adds to make 2?
There are no real numbers that will answer this question, just imaginary numbers. The imaginary numbers are 1+sqrt(-2) and 1-sqrt(-2), which we normally write as 1+2i and 1-2i where i stands for the sort of (-1) (1+2i) + (1-2i) =2 (1+2i) * (1-2i) = 3 (note * stands for multiply)
What is the square-root of 4?
The square root of -4 is 2i: 2i x 2i = 4i2 = 4(-1) = -4
What is the answer to (-3 plus 4i)(8-2i)?
There is a mathematical symbol missing and I am assuming the question to be -2i(8i + 4) - 8(4 - 4i) which is -16i^2 - 8 - 32 - 32i which is -16(-1) - 8 - 32 - 32i which is +16 - 8 - 32 - 32i which is -24 -32i
What is the answer if you Find all the roots of the equation x4 - 2x3 plus 14x2 - 18x plus 45 0 given that 1 plus 2i is one of its roots.?
The four roots are:1 + 2i, 1 - 2i, 3i and -3i.
What is the solution set of the equation x2 minus 2x plus 5 equals 0?
Using the quadratic formula, you get the complex answers of 1 + 2i and 1 - 2i
How do you solve X squared plus 2x plus 2?
x2 +2x+2=0 cannot be factored.If we look at b2 -4ac which is thewe see it is 4-4(2)=4-8 which is negative.This means there two roots and they are not real.The quadratic equation tells us the solution (-2+ 2i)/2 and (-2-2i)/2or -1+i and -1-i.
What is the answer to -2i(8i 4)-8(4-4i)?
There is a mathematical symbol missing and I am assuming the question to be -2i(8i + 4) - 8(4 - 4i) which is -16i^2 - 8 - 32 - 32i which is -16(-1) - 8 - 32 - 32i which is +16 - 8 - 32 - 32i which is -24 -32i
What are the zeros of this polynomial function f of x equals x3 minus 3x2 minus 5x plus 39?
f(x)=x3-3x2-5x+39=(x+3)(x2-6x+13) It has three roots. One of which is x=-3. Using the quadratic equation: x = (6 +/- √(-16))/2 x = (6 +/- 4i)/2 = (3 +/- 2i) so, x=-3, x=3+2i, or x=3-2i
If 3 - 2i is a solution to a polynomial equation is the complex conjugate 3 plus 2i also a solution yes or no?
Not necessarily, take for example the equation x^2=5-12i. Then, 3-2i satisfies the equation. However, 3+2i does not because (3+2i)^2 = 5+12i.
X2 plus 5 equals -2x?
If x2 + 5 = -2x, then x2 + 2x + 5 = 0 This equation cannot be readily factored so using the quadratic formula :- x= {-2 ± √[(-2)2 - 4x5]} ÷ 2 = {-2 ±√-16} ÷ 2 = -1 ± 2i Therefore x = -1 + 2i or x = -1 - 2i
What is the net ionic equation of 2h so42 ca2 2i caso4 2h 2i?
The net ionic equation is SO42- + Ca2+ CaSO4.
What is the net ionic equation of 2h so42- ca2 2i- caso4 2h 2i-?
The net ionic equation is SO42- + Ca2+ CaSO4.
Find 2 square roots of 4i?
2i
How many fourth roots will a positive real number have?
In general, the answer is 4, but only 2 of them are real. For example, the 4th roots of 16 are 2, -2, 2i, and -2i.
