Question

14) $n^2 + n + 1$, find $\frac{1}{\alpha} + \frac{1}{\beta}$ $n^2 + (1+1)n + 1 = 0$ $n^2 + n + 1 = 0$ $n(n+1)n(n+1) = 0$ n = -1, n = -1 \alpha = -1, \beta = -1 $\frac{1}{\alpha} + \frac{1}{\beta}$, $\frac{1}{-1} + \frac{1}{-1}$ $\frac{1+1}{-1} = \frac{2}{-1} = -2$

          14) $n^2 + n + 1$, find $\frac{1}{\alpha} + \frac{1}{\beta}$
$n^2 + (1+1)n + 1 = 0$
$n^2 + n + 1 = 0$
$n(n+1)n(n+1) = 0$
n = -1, n = -1
\alpha = -1, \beta = -1
$\frac{1}{\alpha} + \frac{1}{\beta}$, $\frac{1}{-1} + \frac{1}{-1}$
$\frac{1+1}{-1} = \frac{2}{-1} = -2$

14) n^2 + n + 1, find (1)/(α) + (1)/(β)
n^2 + (1+1)n + 1 = 0
n^2 + n + 1 = 0
n(n+1)n(n+1) = 0
n = -1, n = -1
α= -1, β= -1
(1)/(α) + (1)/(β), (1)/(-1) + (1)/(-1)
(1+1)/(-1) = (2)/(-1) = -2

Added by Lisa C.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

Instant Answer

Solved by Expert Georgiann Andersen

Step 1

The quadratic formula states that for any quadratic equation in the form ax^2 + bx + c = 0, the solutions (or zeros) can be found using the formula: x = (-b ± √(b^2 - 4ac)) / (2a) In our case, a = 1, b = 1, and c = 1. Plugging these values into the quadratic Show more…

Show all steps