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

1 Answer

Answer:

#x_1=(1+sqrt(41))/4#
#x_2=(1-sqrt(41))/4#

Explanation:

From the given #2x^2-x-5=0#

Factor out the 2 from the first two terms so that the coefficient of #x^2# is 1

#2(x^2-1/2x)-5=0#

Now take note of the 1/2 coefficient of x. Divide this 1/2 by 2 it becomes 1/4. Square it, it will become 1/16. This 1/16 will be added and subtracted to the terms inside the grouping symbol.

Let us continue

#2(x^2-1/2x)-5=0#

#2(x^2-1/2x+1/16-1/16)-5=0#

Notice #x^2-1/2x+1/16# is a PST-Perfect Square Trinomial and
#x^2-1/2x+1/16=(x-1/4)^2#

so that

#2((x^2-1/2x+1/16)-1/16)-5=0#

#2((x-1/4)^2-1/16)-5=0#

Now, transpose the 5 to the right side then divide both sides by 2 then transpose the 1/16

#2((x-1/4)^2-1/16)-5=0#

#2((x-1/4)^2-1/16)=5#

#(cancel2((x-1/4)^2-1/16))/cancel2=5/2#

#(x-1/4)^2-1/16=5/2#

#(x-1/4)^2=5/2+1/16#

Simplify

#(x-1/4)^2=41/16#

Extract the square root of both sides

#sqrt((x-1/4)^2)=+-sqrt(41/16)#

#x-1/4=+-1/4sqrt(41)#

#x=1/4+-1/4sqrt(41)#

#x=(1+-sqrt(41))/4#

#x_1=(1+sqrt(41))/4#
#x_2=(1-sqrt(41))/4#

God bless...I hope the explanation is useful.