Question

При каком значении переменной a корень уравнения x(a^2+3)=9x-a будет равен 1? Показать ход решения.

Answer 1

Т.к. корень - x и он должен быть равен 1, то заменим все x на 1:
$1( {a}^{2} + 3) = 9 \times 1 - a$
Решим уравнение:
$a^{2}+a-6 = 0$
[ $a = -3$
[ $a = 2$
Ответ: при a = -3 или 2, x = 1.

Answer 2

1*(a^2+3)=9*1-a

a^2+3=9-a

a^2+a-6=0
D=1-4*1*(-6)=1+24=25=5^2

a1=(-1+5)/2*1=4/2=2

a2=(-1-5)/2*1=-6/2=-3

Answer 3

To find the value of the variable "a" when the root of the equation is equal to 1, we can follow these steps:

Step 1: Start with the given equation: x(a^2 + 3) = 9x - a

Step 2: Set the equation equal to 1, as the root of the equation is given to be 1: x(a^2 + 3) = 1

Step 3: Now, solve for "a" by isolating it on one side of the equation. First, let's distribute "x" on the left side of the equation: a^2x + 3x = 1

Step 4: Rearrange the equation to form a quadratic equation in terms of "a": a^2x - 1 = -3x

Step 5: Now, we can express "x" in terms of "a" using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a

In our case, the coefficients are: a = a^2, b = -1, c = -3x

Step 6: Substitute these values into the quadratic formula for "x": x = (1 ± √((-1)^2 - 4 * a^2 * (-3x))) / 2 * a^2

Step 7: Simplify the expression inside the square root: x = (1 ± √(1 + 12ax)) / 2 * a^2

Step 8: To have a real solution, the expression inside the square root must be greater than or equal to 0: 1 + 12ax ≥ 0

Step 9: Solve for "a": 12ax ≥ -1 a ≥ -1 / (12x)

Now, we have the condition that "a" must be greater than or equal to -1 / (12x). This will ensure that the root of the equation is equal to 1.

Note: It's essential to consider the domain of "x" when solving such equations, as there might be additional restrictions to ensure real solutions. But based on the given equation and condition, this is the value of "a" that satisfies the requirement.