Solve by completing the square. 6 {b}^(2) + 12b - 14 = 0 Please show all work! ​

Question

Solve by completing the square.


`6 {b}^(2) + 12b - 14 = 0`
Please show all work! ​

Answer 1

x = -1 ± √(10/3) = (-3 ± √30)/3

Explanation:

Steps to complete this process:

ax² + bx + c = 0

• Divide by the co-efficient of x² i.e. 'a'

=> (ax² + bx + c)/a = 0/a

=> x² + (b/a)x + (c/a) = 0

=> x² + 2x(b/2a) + (c/a) = 0

Now it seems somewhat like a square (x² + 2x(b/2a), in order to complete:

• Add (b/2a)² to both sides

=> x² + 2(b/2a) + (b/2a)² + (c/a) = (b/2a)²

=> (x + b/2a)² + (c/a) = b²/4a²

=> (x + b/2a)² = (b² - 4ac)/4a²

So on & you can derive quadratic formula.

In the given question:

=> 6b² + 12b - 14 = 0

=> (6b² + 12b - 14)/6 = 0/6

=> b² + 2b - 7/3 = 0

=> b² + 2(1)b + 1² - (7/3) = 1²

=> (b + 1)² - (7/3) = 1

=> (b + 1)² = 10/3 or 30/9

=> b + 1 = ±√(30/9)

=> b = -1 ± √(30)/3

=> b = (-3 ± √30)/3

*this is the rationalized form(rational-denominator), exact answer:

=> (b + 1)² - (7/3) = 1

=> (b + 1) = ± √(10/3)

=> b = -1 ± √(10/3)

You can take LCM, and even rationalize it to (-3 ± √30)/3.

Answer 2

b = - 1 ±
`\sqrt{(10)/(3) }`

Explanation:

Given

6b² + 12b - 14 = 0 ( add 14 to both sides )

6b² + 12b = 14

To complete the square the coefficient of the b² term must be 1

Factor out 6 from the 2 terms on the left side

6(b² + 2b) = 14

To complete the square

add/subtract ( half the coefficient of the b- term)² to b² + 2b

6(b² + 2(1)b + 1 - 1) = 14

6(b + 1)² + 6(- 1) = 14

6(b + 1)² - 6 = 14 ( add 6 to both sides )

6(b + 1)² = 20 ( divide both sides by 6 )

(b + 1)² =
`(20)/(6)` =
`(10)/(3)` ( take the square root of both sides )

b + 1 = ±
`\sqrt{(10)/(3) }` ( subtract 1 from both sides )

b = - 1 ±
`\sqrt{(10)/(3) }` ← exact solution