show x=2-3i is a solution of x^2-4x+13=0

Question

Don't know how to solve this one either

Answer 1

 Can use quadratic formula which derives from factorization of quadratic: aX^2 + bX + c, by completing

 

   the square.

 

 X = -b/2a ±√( b^2 - 4ac)/ 2a

 

      X^2 - 4X + 13 = 0

 

          

 

      X = 4/ 2 ± √( 4^2 - 4(1)(13) / 2

 

       X = 2 ±√-36/ 2

 

           X = 2 ±6i/2

 

            X = 2 ± 3i

 

 

 

Answer 2

Hi Shyanna;

I USED A DIFFERENT APPROACH THAN STEVE OR BRAD AND HAD A DIFFERENT RESULT THAN x=2-3i.  PLEASE CONSIDER IT ANYWAY.

x²-4x+13=0

Let's factor the (x²-4)

(x-2)(x-2)

Let's FOIL...

FIRST...(x)(x)=x²

OUTER...(x)(-2)=-2x

INNER...(-2)(x)=-2x

LAST...(-2)(-2)=4

x²-2x-2x+4

x²-4x+4

Instead of +4, we need +13. (4+9)=13...

x²-4x+4+9=0

(x-2)²+9=0

Let's subtract 9 from both sides...

(x-2)²=-9

Let's square-root both sides...

x-2=√-9

x-2=3i

Let's add 2 to both sides...

x=2+3i

 

Answer 3

An alternative using Nested Form (similar to Synthetic Substitution):

x^2 - 4x + 13
= (x - 4)x + 13

Substituting 2-3i for x:

((2-3i) - 4)(2-3i) + 13
= (2 - 4 - 3i)(2-3i) + 13
= (-2 - 3i)(2-3i) + 13
= -2(2-3i) - 3i(2-3i) + 13
= -4 + 6i - 6i + 9i^2 + 13
= -4 + 9(-1) + 13
= -4 - 9 + 13
= 0

Answer 4

Hey Shyanna -- we may "plug in" [2-3i][2-3i] minus 4[2-3i] plus 13 = 0 ...

 

2*2 minus 6i minus 6i plus 9i*i -8 plus 12i plus 13 ... i-terms cancel [12 less 6 less 6] ...

 

real #'s cancel [4 and 13 -8 less 9] since i*i is -1 ... the key? "signage" ... Regards :)