How do you solve #4x^2+8x+3=0# by completing the square?
2 Answers
Explanation:
Answer:
Explanation:
Step 1. Factor out any numbers in front of the
Step 2. Cut the term in front of
 Term in front of
#x# : 2  Cut it in half:
#1#  Square it:
#1^2=1#  Add and subtract that inside the parenthesis
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)^21+3/4)=0# 
Simplify the remaining addition / subtraction term
#4((x+1)^21/4)=0#
Step 4. Multiply the coefficient term back through
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: