Complex number question?

Question

The question was phrased: Determine whether "1+9i" is the solution of x^2 + 10x + 10 = 0.

How do I determine this, please?

Answer 1

You can either factor it (quadratic formula or complete the square) or just plug in the value and see if it checks.

Factor by completing the square:
x² + 10x + 10 = 0
x² + 10x = -10
x² + 10x + (10/2)² = (10/2)² - 10
(x + 10/2)² = 15
x + 5 = ±sqrt15
x = -5 ± sqrt15

x² + 10x + 10 = (x + 5 + sqrt15)(x + 5 - sqrt15)
1 + 9i is not a root, so it is not a solution.

Plug in the value:
(1 + 9i)² + 10(1 + 9i) + 10
(1 + 18i + 81i²) + (10 + 90i) + 10
1 + 18i - 81 + 10 + 90i + 10
-60 + 108i
The answer is not zero, so 1 + 9i is not a solution.

Answer 2

Let's use the quadratic formula:
x = ½(-10 + √100-40) = ½(-10 + √60)
and
x = ½(-10 - √60).
Since both these roots are real, 1 + 9i is
not a solution of your equation.

Another way:
Plug in 1+9i and see if it satisfies the equation:
(1+9i)² = 1-81 + 18i = -80 + 18i
10(1+9i) = 10 + 90i
10 = 10
Sum: -60 +108i.
Since this isn't zero, 1+9i is not a root.

Third and simplest way:
Compute the discriminant b²-4ac = 100-40 = 60.
This is positive, so both roots are real.
Thus 1+9i is not a root.
Hope that helps!

Answer 3

To determine if something is a solution you can either:

1) Find the solutions yourself and see if it matches any of them or....
2) Plug it back into the equation and see if the equality is true.

I'll proceed with option #2.

If x = 1 + 9i, then
x^2 + 10x + 10
= (1+9i)^2 + 10*(1+9i) + 10
= (1*1+9i+9i+81i^2)+10+90i+10
= 1+18i-81+20+90i
= 108i-60

That does not equal zero so it is NOT a solution to your equation.

Answer 4

Consider substituting x = 1 + 9i into the equation. Considering only the imaginary part of each term.

Imaginary part of (1 + 9i)^2 = 18i
Imaginary part of 10(1 + 9i) = 90i
Imaginary part of 10 = 0

Therefore the expression is not zero for the given value of x.

The real parts do not add up to zero either, as two other answerers have already said, but that takes longer.

Answer 5

1) First method:
You must solve equation x^2 +10x + 10 = 0

Use quadratic formula => x = (-10 ± sqrt(100 -(4. 1.10))/2

= (-10 ± sqrt(100 -60))/2

= (-10 ± sqrt(40))/2

x1 = -1,838

x2 = -8.162

=> Conclusion :2 real roots

2) Second method (shortcut method)

x^2 + 10x +10 = 0

you take sqrt(b^2 -4ac) with b^2 -4ac as discriminant of
quadratic equation.

1) If b^2 -4ac > 0 => (2 real roots)
2) If b^2 -4ac < 0 No, real roots => ( 2 complex roots)
3) If b^2 -4ac = 0 => ( 1 double real roots)

We have : b^2 -4ac = 10^2 -4.1.10 = 100 -40 =60 >0

So, we say that we will obtain 2 real roots.

Answer : ''1 +9i'' is not a solution of x^2 +10x +10 =0

Answer 6

Easier way: the discriminant of this quadratic is
b² - 4ac = 10² - 4*1*10 = 60.

Positive discriminant → two real roots → no imaginary roots → 1+9i is not a root! End of story.