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

2 Answers
May 24, 2017

Answer:

#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#

May 24, 2017

Answer:

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#