Question

Given x square + x +1 =0 find the value of x+1byx

Answer 1

Answer:

Given x

2

+x−(a+2)(a+1)=0, Here, a = 1, b = 1, c = -(a + 2)(a + 1)

x=

2a

−b±

b

2

−4ac

=

2(1)

−1±

1

2

−4{(a+2)(a+1)}

=

2

−1±

1−4{−(a

2

+3a+2)

x=

2

−1±

1+4a

2

+12a+8

=

2

−1±

4a

2

+12a+9

=

2

−1±

(2a+3)

2

=

2

−1±(2a+3)

∴x=

2

−1+2a+3

and x=

2

−1−2a−3

∴x=

2

2a+2

and x=

2

−2a−4

x=a+1x=−(a+2)

∴ The roots of x

2

+x−(a+2)(a+1)=0 are (a+1) and −(a+2)

Answer 2

Answer:

$x + \frac{1}{x} = - 1$

Step-by-step explanation:

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

$x + \frac{1}{x} = \frac{ {x}^{2} + 1 }{x}$

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