Marzo 24, 2020

di Clarice

Come valutate e risolvete $x^{2} + 3 x + 1 = 0$ $x ^ 2 + 3x + 1 = 0$ ?

Risposta:

$x = - \frac{3}{2} + \frac{\sqrt{5}}{2}$ $x = -3/2+sqrt(5)/2$ or $x = - \frac{3}{2} - \frac{\sqrt{5}}{2}$ $x = -3/2-sqrt(5)/2$

Spiegazione:

Usa la differenza di identità dei quadrati:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2=(a-b)(a+b)$

con i $a = (x + \frac{3}{2})$ $a=(x+3/2)$ e $b = \frac{\sqrt{5}}{2}$ $b=sqrt(5)/2$ come segue:

$0 = x^{2} + 3 x + 1$ $0 = x^2+3x+1$

$= {(x + \frac{3}{2})}^{2} - \frac{3^{2}}{2^{2}} + 1$ $= (x+3/2)^2-3^2/2^2+1$

$= {(x + \frac{3}{2})}^{2} - \frac{5}{2^{2}}$ $= (x+3/2)^2-5/2^2$

$= {(x + \frac{3}{2})}^{2} - {(\frac{\sqrt{5}}{2})}^{2}$ $= (x+3/2)^2-(sqrt(5)/2)^2$

$= ((x + \frac{3}{2}) - \frac{\sqrt{5}}{2}) ((x + \frac{3}{2}) + \frac{\sqrt{5}}{2})$ $= ((x+3/2)-sqrt(5)/2)((x+3/2)+sqrt(5)/2)$

$= (x + \frac{3}{2} - \frac{\sqrt{5}}{2}) (x + \frac{3}{2} + \frac{\sqrt{5}}{2})$ $= (x+3/2-sqrt(5)/2)(x+3/2+sqrt(5)/2)$

Quindi:

$x = - \frac{3}{2} + \frac{\sqrt{5}}{2}$ $x = -3/2+sqrt(5)/2$ or $x = - \frac{3}{2} - \frac{\sqrt{5}}{2}$ $x = -3/2-sqrt(5)/2$

Lascia un commento Annulla risposta