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

Question

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

1 Answer

Impact of this question

1373 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-20T22:19:17

See a solution process below:

Explanation:

We can use the quadratic equation to solve this problem:

The quadratic formula states:

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

$x = \frac{- b \pm \sqrt{b^{2} - (4 a c)}}{2 \cdot a}$ $x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))$

Substituting:

$6$ $color(red)(6)$ for $a$ $color(red)(a)$

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

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

$b = \frac{- - 1 \pm \sqrt{{- 1}^{2} - (4 \cdot 6 \cdot 1)}}{2 \cdot 6}$ $b = (-color(blue)(-1) +- sqrt(color(blue)(-1)^2 - (4 * color(red)(6) * color(green)(1))))/(2 * color(red)(6))$

$b = \frac{1 \pm \sqrt{1 - 24}}{12}$ $b = (color(blue)(1) +- sqrt(1 - 24))/12$

$b = \frac{1 \pm \sqrt{- 23}}{12}$ $b = (color(blue)(1) +- sqrt(-23))/12$

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you solve 6b2−b+1=06b ^ { 2} - b + 1= 0?

1 Answer

Explanation:

Related questions

Impact of this question

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