How do you solve #6b ^ { 2} - b + 1= 0#?

1 Answer
Mar 20, 2018

Answer:

See a solution process below:

Explanation:

We can use the quadratic equation to solve this problem:

The quadratic formula states:

For #color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0#, the values of #x# which are the solutions to the equation are given by:

#x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))#

Substituting:

#color(red)(6)# for #color(red)(a)#

#color(blue)(-1)# for #color(blue)(b)#

#color(green)(1)# for #color(green)(c)# gives:

#b = (-color(blue)(-1) +- sqrt(color(blue)(-1)^2 - (4 * color(red)(6) * color(green)(1))))/(2 * color(red)(6))#

#b = (color(blue)(1) +- sqrt(1 - 24))/12#

#b = (color(blue)(1) +- sqrt(-23))/12#