Question

X²-3x-18=0 теорема Виета​

Answer 1

$x^{2}-3x-18=0\\\\x_1+x_2=-\dfrac{b}{a}\;\;\;\;\;\Rightarrow \;\;\;\;\;x_1+x_2=-\dfrac{-3}{1} = 3\\\\x_1\cdot x_2 = \dfrac{c}{a} \; \; \; \; \;\Rightarrow\;\;\;\;\;x_1\cdot x_2=\dfrac{-18}{1}=-18\\\\x_1=-3\\x_2=6$

Ответ: $x_1=-3;\;\;\;x_2=6$

Answer 2

х²–3х–18=0

х¹+х²=3;

х¹×х²=–18.

Answer 3

The Vieta's formulas, also known as Vieta's theorem, provide relationships between the coefficients and roots of a polynomial equation. For a quadratic equation in the form of $ax^2 + bx + c = 0$, the Vieta's formulas state the following:

Let the roots of the equation be $x_1$ and $x_2$. Then:

1. The sum of the roots is given by: $x_1 + x_2 = -\frac{b}{a}$.
2. The product of the roots is given by: $x_1 \times x_2 = \frac{c}{a}$.

In your case, the quadratic equation is $x^2 - 3x - 18 = 0$, where $a = 1$, $b = -3$, and $c = -18$.

Applying the Vieta's formulas:

1. Sum of the roots: $x_1 + x_2 = -\frac{b}{a} = -\frac{-3}{1} = 3$.
2. Product of the roots: $x_1 \times x_2 = \frac{c}{a} = \frac{-18}{1} = -18$.

So, for this quadratic equation, the sum of the roots is 3, and the product of the roots is -18.