How do you solve using the completing the square method #x^2 - 1/2 x - 3/16 = 0#?

1 Answer
Sep 19, 2016

Answer:

#x = 3/4 " or " x= -1/4#

Explanation:

"Completing the Square".

This involves adding the correct value to a quadratic expression to create a perfect square.

Recall: #(x-5)^2 = x^2 color(tomato)(-10)xcolor(tomato)(+25)" "larr color(tomato)(((-10)/2)^2 = +25)#

This relationship between #color(tomato)(b and c)# will always exist.

If the value of #c# is not the correct one, add on what you need.
However, if you have an equation add to BOTH sides.

#x^2 - 1/2 x - 3/16 = 0" " larr# check: is #((-1)/2 div 2)^2 = -3/16?# No

(the last term MUST be positive, because it is squared)

#x^2 - 1/2 x color(blue)(- 3/16) = 0" "larr# not what we want, move it to RHS

#x^2 color(magenta)(- 1/2) x " +? " = color(blue)(3/16) "larr# add on what you DO want

#[color(magenta)(-1/2) rarr div 2 rarr-1/4 rarr "squared" =color(red)(+1/16)]#

#x^2 - 1/2 x color(red)(+1/16) = 3/16 color(red)(+1/16) "larr# add to BOTH sides

Now write the square of the binomial as #(x +?)^2#

#(xcolor(magenta)(-1/4))^2 = 4/16 = 1/4" "larrcolor(magenta)(-1/2 div 2 = -1/4)#

Now solve the equation - isolate #x#

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

#x-1/4 = +-sqrt(1/4)" "larr# square root both sides

#x = +sqrt(1/4) +1/4 = 1/2+1/4 = 3/4#

Or

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