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.
(x - u)*(x - u)*(x + 2i)*(x - 2i) = (x2 - 2xu + u2)*(x2 + 4) = x4 - 2x3u + x2(u2 + 4) - 8xu + 4u2
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
7 + 2i
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)
The square root of -4 is 2i: 2i x 2i = 4i2 = 4(-1) = -4
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
The four roots are:1 + 2i, 1 - 2i, 3i and -3i.
Using the quadratic formula, you get the complex answers of 1 + 2i and 1 - 2i
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.
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
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
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.
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
2i
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.
2H+ + SO42- + Ca2+ + 2I- CaSO4 + 2H+ + 2I
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.