Respuesta :
Answer:
b = - 1 ± [tex]\sqrt{\frac{10}{3} }[/tex]
Step-by-step 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 = [tex]\frac{20}{6}[/tex] = [tex]\frac{10}{3}[/tex] ( take the square root of both sides )
b + 1 = ± [tex]\sqrt{\frac{10}{3} }[/tex] ( subtract 1 from both sides )
b = - 1 ± [tex]\sqrt{\frac{10}{3} }[/tex] ← exact solution
Answer:
x = -1 ± √(10/3) = (-3 ± √30)/3
Step-by-step 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.