Quadratic equations. Vieta's formulas and examples

The solutions $x_1$ and $x_2$ of the quadratic equation $ax^2+bx+c=0$ can be found by special Vieta’s formulas.

$$x_1 + x_2 = -\frac{b}{a}, \ \ x_1 \cdot x_2 = \frac{c}{a}$$

Why use these formulas when we have discriminant formulas? Sometimes to solve equation with discriminant formula can be a bit complicated. Because the solutions of equation can be irrational numbers and other actions wouldn't be easy. So then it is recommended to use Vieta's formulas.

Solved examples

Let's solve some example tasks to see when is it recommended to use these formulas.

Let we have quadratic equation $2x^2-7x+4=0$, and $\{x_{1},x_{2}\}$ are solutions of this equation. Find the $x_{1}+x_{2}$ value.

$$x_{1}+x_{2}=-\frac{b}{a}=-\frac{7}{2}=3.5$$

Let we have quadratic equation $x^2-3x-5=0$, and $\{x_{1},x_{2}\}$ are solutions of this equation. Find the $x_{1}^2+x_{2}^2$ value.

$$x_1 + x_2 = 3, \ \ x_1 \cdot x_2 = -5$$ $$x_{1}^2+x_{2}^2=(x_{1}+x_{2})^2-2x_{1}x_{2}=3^2-2\cdot (-5)=19$$

Let we have quadratic equation $x^2+2x-7=0$, and $\{x_{1},x_{2}\}$ are solutions of this equation. Find the $x_{1}^3+x_{2}^3$ value.

$$x_1 + x_2 = -2, \ \ x_1 \cdot x_2 = -7$$ $$x_{1}^3+x_{2}^3=(x_{1}+x_{2})^3-3x_{1}^2x_{2}-3x_{1}x_{2}^2=(x_{1}+x_{2})^3-3x_{1}x_{2}(x_{1}+x_{2})=(-2)^3-3\cdot (-7)\cdot (-2)=-50$$

2021-07-27