Question

При каких значениях "b" уравнение Х*+bx+25=0 имеет ровно один корень?Для каждого значения "b" укажите соответствующий корень уравнения.Распишите все решение подробно.Заранее большое спасибо;)

Answer 1

X² +bx +25 =0;
уравнение x²+bx+25=0 имеет ровно один корень ,если дискриминант равно нулю :
D = b² - 4*1*25 = 0 ⇔b² =10²⇒[ b= -10; b=10.
a) b= -10;
x²  - 10x +25 =0;
(x-5)² =0⇒x =5;
b) b= 10;
x²  +10x +25 =0;
(x+5)² =0⇒x = -5

Answer 2

Equation with One Root for Different Values of "b"

To determine the values of "b" for which the equation "x * bx + 25 = 0" has exactly one root, we can use the discriminant of the quadratic equation. The discriminant is given by the formula:

D = b^2 - 4ac

In this case, the quadratic equation is in the form "ax^2 + bx + c = 0", with a = 1, b = bx, and c = 25. To have exactly one root, the discriminant must be equal to zero (D = 0).

Let's substitute the values into the discriminant formula and solve for "b":

D = (bx)^2 - 4(1)(25) = 0

Simplifying the equation:

b^2x^2 - 100 = 0

Now, we can solve for "b" by factoring the equation:

b^2x^2 = 100

Taking the square root of both sides:

b * x = ±10

Solving for "b":

b = ±10 / x

Therefore, the equation "x * bx + 25 = 0" has exactly one root for any value of "b" when "b" is equal to ±10 divided by "x".

Please note that the specific value of the root of the equation will depend on the value of "x".