Question

Срочно надо! Помогите, пожалуйста, по теореме Виета составить квадратное уравнение, корни которого равны -3 и 1/3. Заранее спасибо!

Answer 1

${x}^{2} + px + q = 0 \\ x1 + x2 = - p \\ x1 \times x2 = q \\ - 3 + \frac{1}{3} = - 2 \frac{2}{3} \\ - 3 \times \frac{1}{3} = - 1 \: \: = > \\ {x}^{2} + 2 \frac{2}{3} x - 1 = 0$
Вроде бы так, но, честно говоря, я не уверена.

Answer 2

Theorem of Vieta and Quadratic Equations

The theorem of Vieta states that for a quadratic equation of the form ax^2 + bx + c = 0, the sum of the roots is equal to -b/a and the product of the roots is equal to c/a.

To create a quadratic equation with roots -3 and 1/3, we can use the theorem of Vieta to find the coefficients of the equation.

Let's denote the roots as r1 = -3 and r2 = 1/3.

According to the theorem of Vieta, the sum of the roots is equal to the negation of the coefficient of the linear term divided by the coefficient of the quadratic term. Therefore, we have:

r1 + r2 = -b/a

Substituting the values of the roots, we get:

-3 + 1/3 = -b/a

Simplifying the equation, we have:

-9/3 + 1/3 = -b/a

-8/3 = -b/a

Multiplying both sides of the equation by -3/a, we get:

8 = 3b/a

Similarly, the product of the roots is equal to the constant term divided by the coefficient of the quadratic term. Therefore, we have:

r1 * r2 = c/a

Substituting the values of the roots, we get:

(-3) * (1/3) = c/a

Simplifying the equation, we have:

-1 = c/a

Now we have two equations:

8 = 3b/a

-1 = c/a

To find the values of a, b, and c, we need one more equation. We can use the fact that the quadratic equation passes through a specific point, such as the origin (0,0).

Substituting the values of x and y into the equation ax^2 + bx + c = 0, we get:

a(0)^2 + b(0) + c = 0

Simplifying the equation, we have:

c = 0

Now we have three equations:

8 = 3b/a

-1 = c/a

c = 0

From the equation c = 0, we can conclude that c is equal to zero.

Substituting c = 0 into the equation -1 = c/a, we get:

-1 = 0/a

Since a cannot be zero, this equation has no solution.

Therefore, it is not possible to find a quadratic equation with roots -3 and 1/3.

I apologize for the confusion, but it is not possible to satisfy the given conditions for a quadratic equation.