Question

√11+3x-x²=x+3 x+3≥0; x≥-3

Answer 1

11+3x-x^2 = (x+3)^2
11+3x-x^2 = x^2+6x+9
2x^2+3x-2 = 0
D = 9+16 = 25
x1,2 = (-3+/-5)/4
x1 = -2
x2 = 1/2

Answer 2

To solve the equation √11 + 3x - x² = x + 3, and considering the constraints x + 3 ≥ 0 and x ≥ -3, we can follow these steps:

1. Rewrite the equation to isolate the radical term: √11 + 3x - x² - x - 3 = 0

2. Combine like terms: 3x - x² - x + (√11 - 3) = 0

3. Rearrange to get a quadratic equation in standard form: -x² + 2x + (√11 - 3) = 0

4. Now, let's solve for x. To do this, we can use the quadratic formula: x = (-b ± √(b² - 4ac)) / 2a

For our equation, a = -1, b = 2, and c = (√11 - 3).

x = (-(2) ± √((2)² - 4(-1)(√11 - 3))) / 2(-1)

x = (-2 ± √(4 + 4√11 - 12)) / -2

x = (-2 ± √(4√11 - 8)) / -2

x = (2 ± 2√(√11 - 2)) / 2

x = 1 ± √(√11 - 2)

Now, we have two potential solutions:

1. x = 1 + √(√11 - 2)
2. x = 1 - √(√11 - 2)

To check which solutions satisfy the given constraints, we need to evaluate them:

For x = 1 + √(√11 - 2): x + 3 = (1 + √(√11 - 2)) + 3 x + 3 = 4 + √(√11 - 2)

Since x + 3 ≥ 0, we get: 4 + √(√11 - 2) ≥ 0

Subtract 4 from both sides: √(√11 - 2) ≥ -4

The square root of a number is always non-negative, so this solution is valid.

For x = 1 - √(√11 - 2): x + 3 = (1 - √(√11 - 2)) + 3 x + 3 = 4 - √(√11 - 2)

Since x + 3 ≥ 0, we get: 4 - √(√11 - 2) ≥ 0

Again, subtract 4 from both sides: -√(√11 - 2) ≥ -4

However, the square root of a number cannot be negative, so this solution is invalid.

Therefore, the only valid solution to the equation, satisfying the given constraints, is: x = 1 + √(√11 - 2)