Question 373447
2x^2+11x-6=0 How do you complete the square of this problem
:
The coefficient of x^2 should be 1, divide thru by 2
x^2 + {{{11/2}}}x - 3 = 0
:
x^2 + {{{11/2}}}x + __ = 3
:
({{{1/2}}}*{{{11/2}}})^2 = {{{121/16}}} will complete the square, add to both sides
:
x^2 + {{{11/2}}}x + {{{121/16}}} = 3 + {{{121/16}}} 
:
x^2 + {{{11/2}}}x + {{{121/16}}} = {{{48/16}}} + {{{121/16}}} 
:
x^2 + {{{11/2}}}x + {{{121/16}}} = {{{169/16}}} 
:
(x + {{{11/4}}})^2 = {{{169/16}}}
:
Find the square root of both sides
x + {{{11/4}}} = +/-{{{sqrt(169/16)}}} 
:
x + {{{11/4}}} = +/-{{{13/4)}}} 
Two solutions
x = {{{-11/4}}} + {{{13/4)}}}
x = {{{2/4}}} = {{{1/2}}}
and
x = {{{-11/4}}} - {{{13/4)}}}
x = {{{-24/4}}} = -6  
:
:
You can confirm this by checking these values for x in the original problem