Question

Find the two number such that their sum is 6 and product is 14 of complex number

Answer 1

hey mate.
here's the solution

Answer 2

Answer:

$(x,y)=(3+\sqrt{5}i,3-\sqrt{5}i)$

$(x,y)=(3-\sqrt{5}i,3+\sqrt{5}i)$                  

Step-by-step explanation:

Given : The two number such that their sum is 6 and product is 14 of complex number.

To find : The numbers ?

Solution :

Let the number be x and y,

Sum of numbers be 6, i.e. $x+y=6$ .....(1)

Product of numbers be 14, i.e. $xy=14$ .....(2)

Solving (1) and (2),

Substitute y from (1) in (2),

$x(6-x)=14$

$6x-x^2=14$

$x^2-6x+14=0$

Apply quadratic formula, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Here, a=1 , b=-6, c=14

$x=\frac{-(-6)\pm\sqrt{(-6)^2-4(1)(14)}}{2(1)}$

$x=\frac{6\pm\sqrt{36-56}}{2}$

$x=\frac{6\pm\sqrt{-20}}{2}$

$x=\frac{6\pm2\sqrt{5}i}{2}$

$x=3\pm\sqrt{5}i$

$x=3+\sqrt{5}i,3-\sqrt{5}i$

Now, Substitute the value of x in (1),

For $x=3+\sqrt{5}i$

$3+\sqrt{5}i+y=6$

$y=3-\sqrt{5}i$

For $x=3-\sqrt{5}i$

$3-\sqrt{5}i+y=6$

$y=3+\sqrt{5}i$

Therefore, The required numbers are

$(x,y)=(3+\sqrt{5}i,3-\sqrt{5}i)$

or $(x,y)=(3-\sqrt{5}i,3+\sqrt{5}i)$