I x2 + 2x I = 15 solve the equation?

Question

please be detailed

Answer 1

We have:

| x² + 2x | = 15

Now, if x² + 2x >= 0, we have:

x² + 2x = 15

If x² + 2x < 0, we have:

-(x² + 2x) = 15 → x² + 2x = -15

So we solve both these equations:

x² + 2x = 15
x² + 2x - 15 = 0
(x + 5)(x - 3) = 0
x = 3, -5

x² + 2x = -15
x² + 2x + 15 = 0
We use the quadratic formula but find that the discriminant = b² - 4ac = 4 - 4(1)(15) < 0 so there are no real answers.

Usually, in equations such as these, we don't include the complex answers; only the real ones suffice.

Thus, the only solutions are x = 3, -5

Answer 2

Assuming x2 means x squared....

x2 + 2x = 15 -----> x2 + 2x - 15 = 0

(x + 5) (x - 3) = 0 -----> Solve for each x

x + 5 = 0 or x - 3 = 0

x = -5 or x = 3

Since you want the absolute value, the answer cannot be negative, therefore x = 3.

*Check
3(sq) + 2*3 = 15 -----> 9 + 6 = 15

Answer 3

Take away the absolute values and work the problem:

x^2 + 2x - 15 = (x + 5)(x - 3) = 0

x = -5 and x = +3 are both solutions.

Now do this: -(x^2 + 2x) = 15
x^2 + 2x + 1 = -15 + 1 = -14 (completing the square)
(x + 1)^2 = -14
x = -1 +/- i sqrt(14)

This is a pair of complex conjugates (imaginary numbers) that are also solutions.

So there are four solutions to your problem.

Answer 4

|x^2 + 2x| = 15

If |y| = 15, then y could be either 15 or -15

so x^2 + 2x could be either 15 or -15

x^2 + 2x = 15
x^2 + 2x - 15 = 0
(x + 5)(x - 3) = 0
x = -5 or 3

Check (-5)^2 + 2(-5) = 25 - 10 = 15
3^2 + 2(3) = 9 + 6 = 15

Now the 2nd part

x^2 + 2x = -15
x^2 + 2x + 15 = 0

Must use the quadratic formula

x = [-2 +- sqrt(2^2 - 4*1*15)]/2*1
x = [-2 +- sqrt(4 - 60)]/2
x = [-2 +- sqrt(-56)]/2
x = [-2 +- 2sqrt(-14)]/2
x = -1 +- sqrt(-14)

These imaginary roots also satisfy the original problem statement.

[-1 + sqrt(-14)]^2 + 2(-1 + sqrt(-14) =
1 -2sqrt(-14) -14 - 2 + 2sqrt(-14) =
1 - 2 - 14 = -15

[-1 - sqrt(-14)]^2 + 2(-1 - sqrt(-14) =
1 + 2sqrt(-14) - 14 - 2 -2sqrt(-14) =
1 - 14 - 2 = -15

Answer 5

|x^2 + 2x| = 15
x^2 + 2x = 15 or x^2 + 2x = -15

x^2 + 2x - 15 = 0, or x^2 + 2x + 15 = 0
(x + 5)(x - 3) = 0, or

x^2 + 2x + 15 = 0
x = (-b ± sqrt(b^2 - 4ac))/(2a)
x = (-2 ± sqrt(2^2 - 4(1)(15)))/(2(1))
x = (-2 ± sqrt(4 - 60))/2
x = (-2 ± sqrt(-4 * 14))/2
x = (-2 ± 2isqrt(14))/2
x = -1 ± isqrt(2)

x = -5, -1 - isqrt(2), 3, or -1 + isqrt(2)

Answer 6

1X2=9 X BEING 4 2X1 =6 X BEING 2.5 SO 1X2=9 AND 2X1=6 AND 6+9 = 15

Answer 7

Is this suppose to be x^(squared)....cause if not it's super easy....
4x = 15

Even if it is it's still super easy.......

Answer 8

do your own homework