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

1 Answer
Mar 10, 2016

Answer:

#" "2(x-1/4)^2-41/8 => "vertex" -> (x,y) -> (1/4,-41/8)#

x-axis and y-axis Intercepts can be found in the normal way

Explanation:

For a more in depth approach have a look at my solution to
http://socratic.org/s/asD9k2Ch. Diferent values but the method is sound.

Write as:#" "2(x^2-color(red)(1/2)x)-5=0#

For the #color(red)(-1/2)" in "-1/2x# apply: #(-1/2)xx(color(red)(-1/2)) = + 1/4#

So the left hand side becomes:

#" "2(x^2-1/4color(green)(x))-5#

Remove the #color(green)(x)#

#" "2(x^(color(magenta)(2))-1/4)-5#

Move the index (power) #color(magenta)(2)# to outside the brackets

#" "2(xcolor(red)(-1/4))^(color(magenta)(2)) -5 #

Square the constant #color(red)((-1/4)^2=+1/16# and subtract twice its value #color(red)(2xx(=1/4)^2=+1/8)#

#" "2(x-1/4)^2-5-1/8#

#" "2(x-1/4)^2-41/8#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~
#x_("vertex") = (-1)xx(-1/4)= +1/4#

#y_("vertex")= -41/8 =-5 1/8#

Tony B