Question

How do you solve `4x^2+8x+3=0` by completing the square?

Answer 1

`x = - 1/2 and x = - 3/2`

Explanation:

`4x^2 + 8x + 3 = 0`
`4x^2 + 8x + 4 = - 3 + 4`
`4(x^2 + 2x + 1) = 1`
`(x + 1)^2 = 1/4`
`x + 1 = +- 1/2`
`x = - 1 +- 1/2`
`x1 = - 1/2`
`x2 = - 3/2`

Answer 2

Answer: `x=-1/2` and `x=-3/2`

Explanation:

Step 1. Factor out any numbers in front of the `x^2` term

`4(x^2+2x+3/4)=0`

Step 2. Cut the term in front of `x` in half, then square it. Add and subtract the result inside the parenthesis

• Term in front of `x`: 2
• Cut it in half: `1`
• Square it: `1^2=1`
• Add and subtract that inside the parenthesis

`4(x^2+2x+1-1+3/4)=0`

Step 3. Isolate and rewrite your perfect square

• Isolate the perfect square
  `4((x^2+2x+1)-1+3/4)=0`

• Rewrite the perfect square
  `4((x+1)^2-1+3/4)=0`

• Simplify the remaining addition / subtraction term
  `4((x+1)^2-1/4)=0`

Step 4. Multiply the coefficient term back through
`4(x+1)^2-1=0`

Step 5. Solve the equation

• Add `1` to both sides
  `4(x+1)^2=1`

• Divide both sides by `4`
  `(x+1)^2=1/4`

• Square root both sides
  `x+1=+-sqrt(1/4)`

• Subtract `1` from both sides
  `x=-1+-1/2`

Answer: `x=-1/2` and `x=-3/2`