Question

Решить 25∧х +2*а*5∧х +а∧2=0

Answer 1

Відповідь:

Пояснення:

Answer 2

$25^x+2a\cdot 5^x+a^2=0\\5^{2x}+2a\cdot 5^x+a^2=0\\5^x=b,b>0\Rightarrow b^2+2ab+a^2=0\\(a+b)^2=0\Rightarrow a+b=0\\a=-b\Rightarrow a=-5^x\Rightarrow x=log_5(-a)$

Answer 3

To solve the equation 25^x + 2a5^x + a^2 = 0, where "a" is a constant, we can treat it as a quadratic equation in terms of 5^x.

Let y = 5^x, then the equation becomes:

25^x + 2a5^x + a^2 = 0

y^2 + 2ay + a^2 = 0

Now, we can solve this quadratic equation for "y":

Using the quadratic formula, y = (-b ± √(b^2 - 4ac)) / 2a, where a = 1, b = 2*a, and c = a^2:

y = (-(2a) ± √((2a)^2 - 41a^2)) / 2*1

y = (-2a ± √(4a^2 - 4*a^2)) / 2

y = (-2*a ± √(0)) / 2

Since the discriminant (the term inside the square root) is zero, we have a single real root for "y":

y = -2*a / 2

y = -a

Now, remember that y = 5^x:

5^x = -a

To solve for "x," take the logarithm base 5 of both sides:

x = log₅(-a)

Keep in mind that the logarithm of a negative number is not defined in the real number system, so the equation has no real solutions for "x" unless "a" is equal to zero (a = 0).

If "a" is zero, the equation becomes:

25^x + 205^x + 0^2 = 0

25^x = 0

However, 25^x can never be zero for any real value of "x," so there are no real solutions for "x" in this case either.

In conclusion, for any real constant "a," the equation 25^x + 2a5^x + a^2 = 0 has no real solutions for "x."