Question

(2а-5)(2а+5)+25-3а²=а²​

Answer 1

$a^{2} = 0 \\ 1 - a = 0 \\ a1 = 0 \: a2 = 1$

Answer 2

0=0 - верное равенство.

Answer 3

To solve the equation (2a - 5)(2a + 5) + 25 - 3a² = a², we can start by simplifying the expression on the left-hand side:

(2a - 5)(2a + 5) + 25 - 3a² = a²

Expanding the binomial (2a - 5)(2a + 5) using the FOIL method, we get:

(4a² - 10a + 10a - 25) + 25 - 3a² = a²

Simplifying further, we have:

(4a² - 3a²) + (10a - 10a) + (25 + 25) - 25 = 0

Combining like terms, we get:

a² + 50 = 0

To isolate the variable a², we subtract 50 from both sides:

a² = -50

Taking the square root of both sides, we find:

a = ±√(-50)

Since the square root of a negative number is not defined in the real number system, this equation has no real solutions.

Therefore, the equation (2a - 5)(2a + 5) + 25 - 3a² = a² has no real solutions.