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Soluzione - Risolvere equazioni quadratiche completando il quadrato

Exact form:

x_{1} = \frac{1}{2} + \frac{\sqrt{33}}{2}

x_1=\frac{1}{2}+\frac{\sqrt{33}}{2}

x_{2} = \frac{1}{2} - \frac{\sqrt{33}}{2}

x_2=\frac{1}{2}-\frac{\sqrt{33}}{2}

Decimal form:

x_{1} = 3, 372

x_1=3,372

x_{2} = - 2, 372

x_2=-2,372

Guarda i passaggi

Spiegazione passo passo

1. Mueve todos los términos al lado izquierdo de la ecuación

$x^{2} - 1 x - 6 = 2$

Sottrai -2 da entrambi i lati:

$x^{2} - 1 x - 6 - 2 = 2 - 2$

Semplifica l'espressione

$x^{2} - 1 x - 8 = 0$

2. Identifica los coeficientes

Utiliza la forma estándar de una ecuación cuadrática, $a x^{2} + b x + c = 0$ , para encontrar los coeficientes de la ecuación:

$x^{2} - 1 x - 8 = 0$

$a = 1$
$b = - 1$
$c = - 8$

3. Mueve la constante al lado derecho de la ecuación y combina

Agrega $8$ a ambos lados de la ecuación:

$x^{2} - 1 x - 8 = 0$

$x^{2} - 1 x - 8 + 8 = 0 + 8$

$x^{2} - 1 x = 8$

4. Completa el cuadrado

Para convertir el lado izquierdo de la ecuación en un trinomio cuadrado perfecto, añade una nueva constante igual a ${(\frac{b}{2})}^{2}$ a la ecuación:

$b = - 1$

${(\frac{b}{2})}^{2} = {(\frac{- 1}{2})}^{2}$

Use the exponents fraction rule ${(\frac{x}{y})}^{2} = \frac{x^{2}}{y^{2}}$

${(\frac{- 1}{2})}^{2} = \frac{- 1^{2}}{2^{2}}$

$\frac{- 1^{2}}{2^{2}} = \frac{1}{4}$

Agrega $\frac{1}{4}$ a ambos lados de la ecuación:

3 passaggi aggiuntivi

$x^{2} - 1 x = 8$

$x^{2} - 1 x + \frac{1}{4} = 8 + \frac{1}{4}$

Converti il numero intero in una frazione:

$x^{2} - 1 x + \frac{1}{4} = \frac{32}{4} + \frac{1}{4}$

Combina le frazioni:

$x^{2} - 1 x + \frac{1}{4} = \frac{(32 + 1)}{4}$

Combina i numeratori:

$x^{2} - 1 x + \frac{1}{4} = \frac{33}{4}$

Ahora que tenemos un trinomio cuadrado perfecto, podemos escribirlo en forma de cuadrado perfecto al añadir la mitad del coeficiente $b$ , $\frac{b}{2}$ :
$b = - 1$

$\frac{b}{2} = \frac{- 1}{2}$

$x^{2} - 1 x + \frac{1}{4} = \frac{33}{4}$

${(x - \frac{1}{2})}^{2} = \frac{33}{4}$

5. Resuelve para $x$

Toma la raíz cuadrada de ambos lados de la ecuación: IMPORTANTE: Al hallar la raíz cuadrada de una constante, obtenemos dos soluciones: positiva y negativa

${(x - \frac{1}{2})}^{2} = \frac{33}{4}$

$\sqrt{{(x - \frac{1}{2})}^{2}} = \sqrt{\frac{33}{4}}$

Cancel out the square and square root on the left side of the equation:

$x - \frac{1}{2} = \pm \sqrt{\frac{33}{4}}$

Aggiungi $\frac{1}{2}$ a entrambi i lati

$x - \frac{1}{2} + \frac{1}{2} = \frac{1}{2} \pm \sqrt{\frac{33}{4}}$

Semplifica il lato sinistro

$x = \frac{1}{2} \pm \sqrt{\frac{33}{4}}$

$x = \frac{1}{2} \pm \frac{\sqrt{33}}{\sqrt{4}}$

$x = \frac{1}{2} \pm \frac{\sqrt{33}}{2}$

$x_{1} = \frac{1}{2} + \frac{\sqrt{33}}{2}$
$x_{2} = \frac{1}{2} - \frac{\sqrt{33}}{2}$

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Perché imparare questo

In their most basic function, quadratic equations define shapes like circles, ellipses and parabolas. These shapes can in turn be used to predict the curve of an object in motion, such as a ball kicked by football player or shot out of a cannon.
When it comes to an object’s movement through space, what better place to start than space itself, with the revolution of planets around the sun in our solar system. The quadratic equation was used to establish that planets’ orbits are elliptical, not circular. Determining the path and speed an object travels through space is possible even after it has come to a stop: the quadratic equation can calculate how fast a vehicle was moving when it crashed. With information like this, the automotive industry can design brakes to prevent collisions in the future. Many industries use the quadratic equation to predict and thus improve their products’ lifespan and safety.

Termini e argomenti

Risoluzione di equazioni di secondo grado completando il quadrato