the roots of a quadratic equation are: r1= 3+2i and r2= 3-2i ...write a quadratic equation w/ roots r1 & r2?

Answer 1

(x-r1)(x-r2)=0 substitute, and expand.

(x - (3+2i))*(x - (3-2i))=0

x^2 -3x -2xi -3x+2xi +(9 + 4)=0

x^2 - 6x +13 = 0

Done!

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Answer 2

For some quadratic equation f(x), we know that if r is a root,
(x - r) is a factor. Thus, we know our two factors to be

f(x) = (x - [3 + 2i]) (x - [3 - 2i])

All we have to do is expand this.

f(x) = x^2 - x[3 - 2i] - x[3 + 2i] + (3 + 2i)(3 - 2i)

It's easy to expand the last two terms, because it's a difference of squares. Also, let's factor out a -x out of the middle terms.

f(x) = x^2 - x[3 - 2i + 3 + 2i] + (9 - 4i^2)

Note that i^2 = -1

f(x) = x^2 - x[6] + (9 - 4(-1))

Now, it's a matter of easy simplification.

f(x) = x^2 - 6x + 13

Answer 3

since roots are r1 and r2, quadratic equation will be (x-r1)(x-r2) = 0

therefore, [ x - (3 + 2i) ][ x - (3 - 2i) ] = 0

x^2 - x(3-2i) - x(3+2i) + (3-2i)(3+2i) = 0
x^2 - x (3-2i + 3+2i) + (9 + 6i - 6i - 4i^2) = 0
x^2 - x(6+0i) + (9 + 4 + 0i) = 0
x^2 - 6x + 13 = 0

q.e.d =)

Answer 4

Well the sum of the roots is 6
and their product is 13,
so your equation is x² -6x + 13 = 0.
If r1 and r2 are the roots of x² + px + q = 0,
then r1+r2 = -p and r1r2 = q.
This is a handy result which helps one solve many problems.

Answer 5

complext root always come in conjugate pairs so if 4 - 3i is a root then so is 4 + 3i => x = 4 ±3i will be a solution to ax^2 + bx + c = 0 => x - 4 = ± 3i now square both sides => (x - 4)^2 = -9 now expand brackets => x^2 - 8x + 16 = -9 => x^2 -8x + 25 = 0 therefor the Answer is 4.

Answer 6

(x - (3 + 2i))(x - (3 - 2i))
x^2 - (3 - 2i)x - (3 + 2i)x + ((3 - 2i)(3 + 2i))
x^2 - 3x + 2ix - 3x - 2ix + (9 - 4i^2)
x^2 - 6x + (9 - 4(-1))
x^2 - 6x + (9 + 4)
x^2 - 6x + 13

Answer 7

(x-3+2i)(x-3-2i)=0
x^2+x(-3-2i)+x(-3+2i)+(-3+2i)(-3-2i)=0
x^2-6x+9-4i^2=0
x^2-6x+13=0