How do you solve using the completing the square method #4x^2 + 4x + 1 = 49 #?

3 Answers
May 15, 2018

Answer:

#x=-4# or #3#.

Explanation:

#4x^2+4x+1=49# can be written as

#(2x)^2+2xx(2x)xx1+1^2-49=0#

or #(2x+1)^2-7^2=0#

and using #a^2-b^2=(a+b)(a-b)# this becomes

#(2x+1+7)(2x+1-7)=0#

or #(2x+8)(2x-6)=0#

and either #2x+8=0# i.e. #2x=-8# and #x=-8/2=-4#

or #2x-6=0# i.e. #2x=6# and #x=6/2=3#

May 15, 2018

Answer:

3, and - 4

Explanation:

  1. Transforming Method
    #y = 4x^2 + 4x - 48 = 0#
    #y = 4(x^2 + x - 12) = 0#
    Solve the quadratic equation #x^2 + x - 12 = 0#, by the new Transforming Method - case a = 1 (Socratic, Google Search)
    Find 2 real roots, that have opposite signs (ac < 0), knowing the sum (-b = -1) and the product (c = -12).
    They are 3 and - 4
  2. Completing the square method.
    #4x^2 + 4x = 48#
    #x^2 + x = 12#
    #x^2 + x + 1/4 = 12 + 1/4 = 49/4#
    #(x + 1/2)^2 = 49/4#
    #x + 1/2 = +- 7/2#
    #x1 = 7/2 - 1/2 = 6/2 = 3#
    #x2 = - 7/2 - 1/2 = - 8/2 = - 4#
May 15, 2018

Answer:

#x=-4 and x= 3#

#4(x+1/2)^2-49=0#

Explanation:

Given: #4x^2+4x+1=49" "...................Equation(1)#

Write as:
#4(x^2+4/4 x)+1-49=0 larr# As yet no values have changed.

The next step will change things in as much as it introduces a value that is not in the original equation. It is removed by introducing an as yet unknown constant of #k#. It will turn the introduced value into 0.

#4(x^(color(red)(2))+ubrace(4/4)color(white)("d") x)+1-49=0#
#color(white)("ddddddd")darr#
#color(white)("dddd")"Halve this."#
#color(white)("ddddddd")darr#
#4(xcolor(white)("d")+color(white)("d")1/2color(white)("d"))^(color(red)(2))+k-48=0" "..................Equation(2)#

Notice the exponent #(color(red)(2))# has been moved outside the bracket, the #x# from #4/4 x# has been removed and that #k# has been introduced. Also as indicated, the #4/4# has been halved as well.

#color(brown)("The above set of steps form the basis for completing the square")#

Now we do the correction bit.

Set #4(1/2)^2+k=0color(white)("ddd")=>color(white)("ddd") 1+k=0color(white)("d")=>k=-1#

Substitute into #Equation(2)# giving:

#4(x+1/2)^2-1-48=0#

#4(x+1/2)^2-49=0 larr #Completed square form
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Add 49 to both sides

#4(x+1/2)^2=49#

Divide both sides by 4

#(x+1/2)^2=49/4#

Square root both sides

#x+1/2=+-7/2#

Subtract #1/2# from both sides

#x=-1/2+-7/2#

#x=-4 and x= 3#

Tony B