SOLUTION: Could you help solve the quadratic equatin by completing the square 2x^2 - 6x + 5 Thank You, Stacie Rihl

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Could you help solve the quadratic equatin by completing the square 2x^2 - 6x + 5 Thank You, Stacie Rihl      Log On


   

Question 260559: Could you help solve the quadratic equatin by completing the square
2x^2 - 6x + 5
Thank You,
Stacie Rihl
Found 2 solutions by onlinetutor365.com, jim_thompson5910:
Answer by onlinetutor365.com(14) About Me  (Show Source):
You can put this solution on YOUR website!

2(x2) - 6x + 5 = 0
Solving for variable 'x'.
Begin completing the square. Divide all terms by 2 the coefficient of the squared term:
Divide each side by '2'.
x^2 - 3x + 2.5 = 0
Move the constant term to the right:
Add '-2.5' to each side of the equation.
x^2-3x +2.5 -2.5=-2.5
x^2 -3x =-2.5

The x term is -3x. Take half its coefficient (-1.5).
Square it (2.25) and add it to both sides.
Add '2.25' to each side of the equation.
x2-3x + 2.25 = -2.5 + 2.25

x2-3x + 2.25= -0.25
Factor a perfect square on the left side:
(x -1.5)(x -1.5) = -0.25
Can't calculate square root of the right side.
The solution to this equation could not be determined.

Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

2x%5E2-6x%2B5 Start with the given expression.


2%28x%5E2-3x%2B5%2F2%29 Factor out the x%5E2 coefficient 2. This step is very important: the x%5E2 coefficient must be equal to 1.


Take half of the x coefficient -3 to get -3%2F2. In other words, %281%2F2%29%28-3%29=-3%2F2.


Now square -3%2F2 to get 9%2F4. In other words, %28-3%2F2%29%5E2=%28-3%2F2%29%28-3%2F2%29=9%2F4


2%28x%5E2-3x%2Bhighlight%289%2F4-9%2F4%29%2B5%2F2%29 Now add and subtract 9%2F4 inside the parenthesis. Make sure to place this after the "x" term. Notice how 9%2F4-9%2F4=0. So the expression is not changed.


2%28%28x%5E2-3x%2B9%2F4%29-9%2F4%2B5%2F2%29 Group the first three terms.


2%28%28x-3%2F2%29%5E2-9%2F4%2B5%2F2%29 Factor x%5E2-3x%2B9%2F4 to get %28x-3%2F2%29%5E2.


2%28%28x-3%2F2%29%5E2%2B1%2F4%29 Combine like terms.


2%28x-3%2F2%29%5E2%2B2%281%2F4%29 Distribute.


2%28x-3%2F2%29%5E2%2B1%2F2 Multiply.


So after completing the square, 2x%5E2-6x%2B5 transforms to 2%28x-3%2F2%29%5E2%2B1%2F2. So 2x%5E2-6x%2B5=2%28x-3%2F2%29%5E2%2B1%2F2.


So 2x%5E2-6x%2B5=0 is equivalent to 2%28x-3%2F2%29%5E2%2B1%2F2=0.


Now let's solve 2%28x-3%2F2%29%5E2%2B1%2F2=0


2%28x-3%2F2%29%5E2%2B1%2F2=0 Start with the given equation.


2%28x-3%2F2%29%5E2=0-1%2F2 Subtract 1%2F2 from both sides.


2%28x-3%2F2%29%5E2=-1%2F2 Combine like terms.


%28x-3%2F2%29%5E2=%28-1%2F2%29%2F%282%29 Divide both sides by 2.


%28x-3%2F2%29%5E2=-1%2F4 Reduce.


x-3%2F2=%22%22%2B-sqrt%28-1%2F4%29 Take the square root of both sides.


x-3%2F2=sqrt%28-1%2F4%29 or x-3%2F2=-sqrt%28-1%2F4%29 Break up the "plus/minus" to form two equations.


x-3%2F2=%281%2F2%29i or x-3%2F2=-%281%2F2%29i Take the square root of -1%2F4 to get %281%2F2%29i.


Note: i=sqrt%28-1%29


x=3%2F2%2B%281%2F2%29i or x=3%2F2-%281%2F2%29i Add 3%2F2 to both sides.


--------------------------------------


Answer:


So the solutions are x=3%2F2%2B%281%2F2%29i or x=3%2F2-%281%2F2%29i.