solve by means of completing the square: 2s(squared)-6s+7=0?

Question

how do u do this by completing the square and pls explain

Answer 1

you first divide by 2
s^2 - 3s + 7/2=0
subtract 7/2 from both sides
s^2 - 3s = -(7/2)
now divide 3/2 and square
s^2 - 3s + 9/4 = 9/4 - 28/4
[s - (3/2)]^2 = -(19/4)
s - 3/2 =( i sqrt 19)/2
s= [3 + or - (i sqrt 19)]/2

Answer 2

1) Make sure the first coefficient is 1
2s^2 - 6s + 7 = 0
s^2 - 3s + 7/2 = 0

2) Move the last term to the other side
s^2 - 3s = -7/2

3) Take half of the middle term, square it, and then add it to both sides
s^2 - 3s + (3/2)^2 = -7/2 + (3/2)^2

4) Now you can factor the left into a perfect square, which lets you solve for s.
(s - 3/2)^2 = -7/2 + (3/2)^2
(s - 3/2)^2 = -7/2 + 9/4
(s - 3/2)^2 = -14/4 + 9/4
(s - 3/2)^2 = -5/4

Normally we could just take the square root of both sides and add the "3/2" to both sides to solve for s. However in this case we have the square of a number on the left equally a negative number. So there is no real solution.

Answer 3

To do a completing a square,
1. Make sure the coeffecient of x^2 is 1. If not, divide the equation throughout by its coeffecient.
2. Next, add a term of
(coeffecient of x/2)^2 - (coeffecient of x/2)^2.
3. Simplify the equation and your equation would be in completed square form.

Here's an example,
2s^2 - 6s + 7 = 0

Dividing throughout by 2,
s^2 - 3s + 3.5 = 0

Adding a term of (coeffecient of x/2)^2 - (coeffecient of x/2)^2,
s^2 - 3s + (-3/2)^2 - (-3/2)^2 + 3.5 = 0

Simplifying the equation,
[s^2 - 3s + (-3/2)^2] + [3.5 - (-1.5)^2] = 0
(s-1.5)^2 + 1.25 = 0

Answer 4

2 S^2 - 6 S + 7 = 0
s^2 - 3 S + 7/2 = 0
s^2 - 3 S + ( 3/2 )^2 - ( 3/2 )^2 + 7/2 = 0
(s + 3/2)^2 - 9 / 4 + 7 / 2 = 0
(s + 3/2)^2 + (- 9 + 14 ) / 4 = 0
(s + 3/2)^2 + 5 / 4 = 0

Answer 5

2(s² - 3s + 7/2) = 0
2.(s² - 3s + 9/4 - 9/4 + 7/2) = 0
s² - 3s + 9/4 = 9/4 - 7/2
(s - 3/2)² = - 5/4
(s - 3/2)² = i² (5/4)
s - 3/2 = ± (i /2) .√5
s = 3/2 ± (i / 2).√5

Answer 6

como best answer.