Write a quadratic equation having the given solutions?

Question

3+4i, 3-4i

Answer 1

so easy :
x^2 - Sx + P = 0

where, S is the sum of the roots and P is the Product of the Roots :

S = 3+4i + 3-4i = 6

P = (3+4i)(3-4i) = 9 - (-16)
= 9+16 = 25

==> the equation is x^2 - 6 + 25 = 0

hope this helps enjoy !!??!!

Answer 2

For a general quadratic eqn :-
px^2+qx +r=0
sum of roots= -(q/p) =6
i.e q = -6p
product of roots = r/p = (3+4i)(3-4i) = 3^2 + 4^2 =25
i.e r = 25p
therefore general eqn satisfying it is
p(x^2 -6x + 25) = 0

Answer 3

the ans is x^2-6x+25.There are 2 ways of doing it:
1) (X-(3-4i))*(X-(3+4I))
= x^2-(3-4i)x-(3+4i)x+(3^2-(4i)^2)
= x^2-6x+(3^2+4^2)
= x^2-6x+25
2) for any quadratic equation ax^2+bx+c,if p &
q are its roots,
Then p+q=-b/a
p*q=c/a
here p= 3+4i
q=3-4i,
then b/a=-((3+4i)+3-4i)
=-6
p*q=(3+4i)*(3-4i)
=3^2+4^2
=25
when a=1
equation=x^2-6x+25

Answer 4

while looking the quadratic equation from given numbers, only subtract them from x (because of the fact the two area of the equation has to equivalent 0: (x + 7)(x - 4) = x^2 + 7x - 4x - 28 = x^2 + 3x - 28

Answer 5

(x - (3 + 4i))(x - (3 - 4i))
x^2 - (3 - 4i)x - (3 + 4i)x + ((3 + 4i)(3 - 4i))
x^2 - 3x + 4ix - 3x - 4ix + (9 - 12i + 12i - 16i^2)
x^2 - 6x + (9 - 16i^2)
x^2 - 6x + (9 - 16(-1))
x^2 - 6x + (9 + 16)
x^2 - 6x + 25

ANS : x^2 - 6x + 25

Answer 6

x^2-6x+25=0

Answer 7

x^2-6x+25=0

Answer 8

(x-3) = +/-4i
or x^-6x+9 = -16
or x^2-6x+25 = 0

Answer 9

(x-3-4i)(x+4i-3)
=x^2+4ix-3x-3x-12i+9-4ix+16+12i
=x^2-6x+25

Answer 10

do ur own dam math hw. (x-(3+4i))(x-(3-4i)