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# 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.

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