Which of the following is one root of the equation x^2-4x+13=0?

Question

the choices are:
1. -1
2. 5
3. 4+ 3i
4. 2- 6i
5. 2+ 3i

Answer 1

x = [ 4 ± √(16 - 52) ] / 2
x = [ 4 ± √(- 36) ] / 2
x = [ 4 ± 6 i ] / 2
x = 2 ± 3i

ANSWER 5

Answer 2

You can either use the quadratic equation to get the solutions or you can sub each possible answer into the equation and see if it is equal to zero.

Using the quadratic equation
x = ( 4±√[4²-4*1*13] ) / 2
=( 4±√[16-52] ) / 2
=( 4±√[-36] ) / 2
=( 4±6i ) / 2
=2±3i

So option 5 is a one of the roots.

Via subbing in:
1) (-1)² - 4*1 + 13 = 18
2) 5² - 4*5 + 13 = 18
3) (4+3i)^2-4(4+3i)+13 = 4+12i
4)(2-6i)^2-4(2-6i)+13 = -27
5) (2+3i)^2-4(2+3i)+13 = 0

So again this way it is option 5.

@Ramesh & Ani - You have both forgotten your brackets in your quadratic equation. Its everything divided by 2.

Answer 3

4 and 5

Answer 4

5. 2+ 3i
as (2 + 3i)^2 -- 4(2 + 3i) + 13 = 4 + 12i -- 9 -- 8 -- 12i + 13 = 0

Answer 5

Use the quadratic formula. You'll get root(-36) as b^2-4ac. This means there must be a 6i in the answer (so it must be 4)

Answer 6

it is 2+3i

eqn is of form :ax^2+bx+c=0
root is (-b±sqrt(b^2-4ac))/2
solving, you get x=2±3i

Answer 7

x=2-sqrt(16-52)/2
2-6i (4) is correct