Giải phương trình

To solve the equation x^2 + (x + |x + 1|) - 14 = 0, we can break it down into cases based on the value of x + 1.

Case 1: x + 1 ≥ 0
In this case, |x + 1| = x + 1, so the equation becomes x^2 + (x + (x + 1)) - 14 = 0.
Simplifying further, we get x^2 + 2x + 1 - 14 = 0.
Combining like terms, we have x^2 + 2x - 13 = 0.

Case 2: x + 1 < 0
In this case, |x + 1| = -(x + 1), so the equation becomes x^2 + (x - (x + 1)) - 14 = 0.
Simplifying further, we get x^2 - 1 - 14 = 0.
Combining like terms, we have x^2 - 15 = 0.

Now, let's solve each case separately:

Case 1: x^2 + 2x - 13 = 0
Using the quadratic formula, x = (-b ± √(b^2 - 4ac)) / (2a), where a = 1, b = 2, and c = -13.
Plugging in these values, we get x = (-2 ± √(2^2 - 4(1)(-13))) / (2(1)).
Simplifying further, we have x = (-2 ± √(4 + 52)) / 2.
This becomes x = (-2 ± √56) / 2.
Simplifying the square root, we have x = (-2 ± 2√14) / 2.
Dividing both the numerator and denominator by 2, we get x = -1 ± √14.

Case 2: x^2 - 15 = 0
Using the quadratic formula, x = (-b ± √(b^2 - 4ac)) / (2a), where a = 1, b = 0, and c = -15.
Plugging in these values, we get x = (0 ± √(0^2 - 4(1)(-15))) / (2(1)).
Simplifying further, we have x = (0 ± √(0 + 60)) / 2.
This becomes x = (0 ± √60) / 2.
Simplifying the square root, we have x = (0 ± 2√15) / 2.
Dividing both the numerator and denominator by 2, we get x = ± √15.

Therefore, the solutions to the equation x^2 + (x + |x + 1|) - 14 = 0 are:
x = -1 + √14, -1 - √14, √15, and -√15.