How do you solve the equation #4x^2+9x-1=0# by completing the square?

2 Answers
Jun 3, 2017

Answer:

Solution: #x=1/8(-9+sqrt97) or x=1/8(-9-sqrt97)#

Explanation:

#4x^2+9x-1 = 4(x^2+9/4x) -1 = 4(x^2+9/4x+ (9/8)^2)-81/16 -1 # or

# 4(x+ 9/8)^2= 81/16 +1 = 4(x+ 9/8)^2= 97/16 = (x+ 9/8)^2= 97/64# or

#(x+ 9/8)= +- sqrt(97/64) or (x+ 9/8)= +- sqrt(97)/8 # or

#:. x = -9/8 +- sqrt(97)/8 or x =1/8(-9+-sqrt97)#

Solution: #x=1/8(-9+sqrt97) or x=1/8(-9-sqrt97)# [Ans]

Jun 3, 2017

Answer:

#x=-9/8+-sqrt(97)/8#

Explanation:

Step 1. Factor out the 4 from in front of the #x^2#

#4(x^2+9/4 x-1/4)=0#

Step 2. Take the coefficient of the #x# term (i.e., #9//4#), cut it in half, square it, and add it and subtract it inside the parenthesis.

When you add a number and subtract it, that's like adding #0# and so doesn't change the problem.

Middle term: #9/4#

Cut it in half: #9/8# (this number will be used in Step 3)

Square the result: #(9/8)^2#

Next, add and subtract this term

#4(x^2+9/4 x" "color(red)( + (9/8)^2)color(red)(-(9/8)^2)-1/4)=0#

Step 3. Use the left three terms to make a perfect square

#4(color(blue)(x^2+9/4 x + (9/8)^2)-(9/8)^2-1/4)=0#

#4(color(blue)((x+9/8)^2)-(9/8)^2-1/4)=0#

Step 4. Simplify the fraction and multiply #4# back through.

#4((x+9/8)^2-(9/8)^2-1/4)=0#

#4((x+9/8)^2-97/64)=0# (*Don't forget signs here like I had!)

#4(x+9/8)^2-97/16=0#

Step 5. Use algebra to solve for #x#

#4(x+9/8)^2=97/16#

#(x+9/8)^2=97/64#

#x+9/8=+-sqrt(97/64)#

#x=-9/8+-sqrt(97)/8#