How do you solve the equation #2x^2+3x-5=0# by completing the square?

1 Answer
Dec 21, 2016

Answer:

#x=-5/2 " and x=1#

Explanation:

Write as #2(x^2+3/2x)-5+k=0.......Equation(1)#

Where #k# is a constant of correction that compensates for the error introduce whilst manipulating the equation.

Take the power outside the bracket.

#2(x+3/2x)^2-5+k=0#

Remove the #x# from #3/2x#

#2(x+3/2)^2-5+k=0#

Halve the #3/2#

#2(x+3/4)^2-5+k=0#

#color(red)("The error comes from")#

#color(red)(2)(x+color(red)(3/4))^(color(red)(2))-5+k=0...Equation(1_a)#
.................................................................................................................
in that we have #color(red)(2xx(3/4)^2)+k=0 larr" building the error correction"#

#=>k=-(2xx9/16) = -18/16 = -9/8#
.....................................................................................................................

#=>2(x+3/4)^2-5-9/8=0#

#=>color(blue)(2(x+3/4)^2-49/8=0)....................Equation(1_b)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#(x+3/4)^2=49/16#

#x+3/4=+-sqrt(49/16)" "=" "+-7/4#

#x=+-7/4-3/4#

#x=-10/4 " and "1#

#color(red)(x=-5/2 " and "1)#

Tony B