
A) -x²-x+20=0 Б) x²+16=0 В) x²-16=0 Г) (x-4)²=0


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Let's solve each quadratic equation:
A) -x² - x + 20 = 0
To solve this quadratic equation, we can use the quadratic formula: x = (-b ± √(b² - 4ac)) / 2a.
For the given equation, a = -1, b = -1, and c = 20.
x = (1 ± √((-1)² - 4 * (-1) * 20)) / (2 * (-1)) x = (1 ± √(1 + 80)) / -2 x = (1 ± √81) / -2 x = (1 ± 9) / -2
So, the solutions are: x₁ = (1 + 9) / -2 = -10 / -2 = 5 x₂ = (1 - 9) / -2 = -8 / -2 = 4
Therefore, the solutions to the equation A) -x² - x + 20 = 0 are x = 5 and x = 4.
B) x² + 16 = 0
To solve this quadratic equation, we can again use the quadratic formula: x = (-b ± √(b² - 4ac)) / 2a.
For the given equation, a = 1, b = 0, and c = 16.
x = (0 ± √(0² - 4 * 1 * 16)) / (2 * 1) x = (0 ± √(0 - 64)) / 2 x = (0 ± √(-64)) / 2
The expression √(-64) represents imaginary numbers since you cannot take the square root of a negative number in the real number system.
Therefore, the equation x² + 16 = 0 has no real solutions, but it has two complex solutions:
x = 0 + 8i (where i is the imaginary unit, i.e., √(-1)) x = 0 - 8i
C) x² - 16 = 0
To solve this quadratic equation, we can use the difference of squares formula: (a² - b²) = (a + b)(a - b).
For the given equation, it can be rewritten as (x + 4)(x - 4) = 0.
Now, we can solve for x:
x + 4 = 0 -> x = -4 x - 4 = 0 -> x = 4
Therefore, the solutions to the equation C) x² - 16 = 0 are x = -4 and x = 4.
D) (x - 4)² = 0
To solve this equation, we can use the fact that (a - b)² = a² - 2ab + b².
For the given equation, it becomes: x² - 8x + 16 = 0.
Now, we can solve for x:
Using the quadratic formula: x = (-b ± √(b² - 4ac)) / 2a a = 1, b = -8, and c = 16.
x = (8 ± √((-8)² - 4 * 1 * 16)) / 2 x = (8 ± √(64 - 64)) / 2 x = (8 ± √0) / 2 x = (8 ± 0) / 2 x = 8 / 2 x = 4
Therefore, the solution to the equation D) (x - 4)² = 0 is x = 4.
In summary: A) The solutions are x = 5 and x = 4. B) The solutions are complex (x = 0 + 8i and x = 0 - 8i). C) The solutions are x = -4 and x = 4. D) The solution is x = 4.


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