Question

Pregunta de algebra, 5to de sec

Answer 1

Ecuación:

x² - (n² - 5)x + (-8n + 3) = 0

Sea "r" la otra raíz.

Suma de raíces:

$\boxed{-3+r=\frac{-(-(n^{2}-5)) }{1} }\\\\\\\boxed{r-3=n^{2}-5 }\\\\\\\boxed{r+2=n^{2} }\\\\\\\boxed{r=n^{2} -2}$

Producto de raíces:

$\boxed{(-3)(r)=\frac{-8n+3}{1} }\\\\\\\boxed{-3r=-8n+3}\\\\\\\boxed{r=\frac{-8n+3}{-3} }\\\\\\\boxed{r=\frac{8n-3}{3} }$

Igualamos:

$\boxed{n^{2}-2=\frac{8n-3}{3} }\\\\\\\boxed{3n^{2} -6=8n-3}\\\\\\\boxed{3n^{2}-8n-3=0 }\\\\\\\boxed{(3n+1)(n-3)=0}\\\\\\\boxed{n=-\frac{1}{3} }\ \boxed{n=3}$

Si n = -1/3

⇒ r = (-1/3)² - 2 = 1/9 - 2 = -17/9

Ecuación:

x² + 44x/9 + 17/3 = 0

Si n = 3

⇒ r = (3)² - 2 = 9 - 2 = 7

Ecuación:

x² - 4x - 21 = 0

∴ La otra raíz puede tener 2 valores: 7 y $-\dfrac{17}{9}$