How can you make a quadratic equation that has solutions of 3+2i and 3-2i?

Question

Also help me how to solve that 3+2i and 3- 2i are in fact solutions.

Answer 1

Sum of roots: 3 + 2i + 3 - 2i = 6
Product of roots: (3+2i)(3-2i) = 9 + 4 = 13

ax² + bx + c = 0
If a=1
Sum of roots = -b
Product of roots = c

Required quadratic:
x² - 6x + 13 = 0

Solve the quadratic using the quadratic formula:
x = (-b±sqrt(b²-4ac))/2a
a = 1
b = -6
c = 13

x = (6±sqrt(6²-4*1*13))/2
= (6±sqrt(36-52))/2
= (6±sqrt(-16))//2
= (6±4i)/2
= 3±2i

Answer 2

The factors of an equation are just (x-a) where a is the roots (solutions) of the equations.

( x - (3+2i) )( x - (3-2i) )
( x - 3-2i )( x - 3+2i )
x² + (-3-2i)x + (-3+2i)x + (-3-2i)(-3+2i)
x² + (-3-2i -3+2i)x + -3*-3 -2i*-3 -3*2i -2i*2i
x² -6x + 9 + 6i - 6i + 4
x² -6x + 13

x² -6x + 13= 0 has roots as required.

Answer 3

first of all, the equation is x = -b +/- sqrt(b^2 - 4ac) / 2a you have: x = -3 +/- 2i sqrt(5) = -3 +/- sqrt(-20) / (2*one million/2) enable b = 3, and a = one million/2, to make the equation artwork. So with b and a we could get: x = -3 +/- sqrt(9 - 4*(one million/2)*c) / (2*one million/2) c could desire to equivalent 29/2 the equation turns right into one million/2*x^2 + 3*x + 29/2 = 0 multiply via 2 x^2 + 6x + 29 = 0

Answer 4

the solutions are (3+2i) and (3-2i)

so...

(x-(3+2i))(x-(3-2i))
(x-3-2i)(x-3+2i)

now multiply out:

x^2 - (3x)- (3x) -2xi + 2xi +9 -4(i^2)
x^2 - 6x +9 + 4

x^2 -6x + 13

Answer 5

[ x - (3 + 2 i) ] [ x - (3 - 2 i) ] = 0
x ² - (3 - 2 i )x - (3 + 2 i )x + (3 + 2 i)(3 - 2i) = 0
x ² + 3x + 2i x - 3x - 2i x + (9 + 4) = 0
x ² + 13 = 0
x ² = ± √ (- 13)
x = ± i √13

Proof
x = [0 ± √(0 - 52) ] / 2
x = [ √ ( - 52 ) ] / 2
x = [ √52 i ² ] / 2
x = (± 2√13 i ) / 2
x = ( ± i √ 13 )

Answer 6

(x-3-2i)(x-3+2i)=x^2-6x+13=y

Answer 7

Given s1 and s2 are solutions to your quadratic, then the expression [x-s1] [x-s2] = 0 is your quadratic eqtn that has those solutions.