math question?

Question

1: factor a^2 + b^2 using complex numbers

2: If the roots are 1+i , 1-i, what is the quadratic equasion?

Plz show work, because i have more questions similar to these. I dont wana post them because i actually wana try.

Answer 1

1. a²+b²
a² - (ib)² since (ib)² = -b²
(√a + i√b)(√a - i√b)

2. If the roots of a quadratic equation are 1+i and 1-i then

x-(1+i)=0
x-(1-i)=0

[x-(1+i)][x-(1-i)]=0

x² - (1+i)x - (1-i)x + (1+i)(1-i) = 0

x² - x - ix - x + ix + 1 + i - i - i² = 0

x² - 2x + 2 = 0

to check we solve this quadratic equation with a=1, b=-2 and c=2

x = [-b ± √(b² - 4ac)]/(2a)
x = [2 ± √(4 - 8)]/2
x = [2 ± √-4]/2
x = [2 ± 2i]/2

x = 1±i

Answer 2

1. this is a "difference" of two squares namely a^2 - i^2*b^2
where i is the square root of -1

factoring this we have
a^2 + b^2 =
a^2 - (-1) b^2 =
a^2 - i^2*b^2
(a + ib)(a - ib)

2. for this, square the imaginary part and put it in the right side of the equation. the left hand side is the square of (x minus the real part)

for this case:
(x-1)^2=(+/- i)^2
(x-1)^2=- 1
x^2 - 2x + 1 = -1
x^2 - 2x + 2 = 0

Answer 3

ur first question is abrupt

second question
we have x = 1 + i and x = 1 - i
so,
x - (1+ i) = 0 and also x - (1 - i) =0
this implies
[x - ( 1 + i)] [x - (1 - i)] = 0
then multiplying we get,
x^2 - x + xi - x + 1 - i - xi + i - i^2 = 0
x^2 - 2x + 2 = 0 [ since i^2 = (-)1]
is the required solution for the given roots

Answer 4

^ = Power and * = multiplication

1) a^2 + b^2 = a^2 + (-1) * (- b^2), but we know the i = sq rt of (-1) and i^2 = -1 = (sq rt of -1)^2 . Therefore
a^2 - (i^2)*(b^2) = (a - ib) * (a+ib)

2) x1 = 1+i and x2 = 1-i >>>

x1 + x2 = S = 2 and P=x1*x2 = 1-i^2 = 1+1=2

The equation is: x^2 - xS +P = 0
x^2 -2x +2 =0 is your desired equation.

Answer 5

(a + b*i)*(a - b*i)
i is an imaginary number: Sq rt of -1
So, when we factor these numbers back in, we get:
a^2 - a*b*i + a*b*i - b*i^2
Simplify:
a^2 - b*i^2
Since i is the square root of -1, i^2 = -1
So, you get a^2 + b^2

For the second question, just do this:
(x - (1+i))*(x - (1-i))
x^2 - x + xi - x - xi - i^2 + 1
therefore, we get: x^2 -2x + 2

Answer 6

a² + b² factors as (a + ib)*(a - ib) (multiply it out just like any other pair of binomials, taking into account the fact that i² = -1)
(a+ib)*(a-ib) = a² + iab - iab - (ib)² = a² - i²b² = a²+b²

Since the roots of a quadratic (or, for that matter, a polynomial of --any-- order) are the x values for which the value of the polynomial goes to zero, just take the product of
(x-(1+i))*(x - (1-i)) and set it equal to zero.

HTH

Doug