How do you solve using the completing the square method #x^2 + 4x = 21#?

1 Answer
Mar 9, 2016

Answer:

#color(blue)("Vertex"->(x,y) = (-2,-25))#

Explanation:

Write as:#" "x^2+4x-21=0#...................(1)

'~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Shortcut method")#

By sight:

#(x+2)^2-4-21=0#

#color(blue)((x+2)^2-25=0)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("The logic behind the shortcut method")#

#color(green)("Consider only the left hand side")#

#color(green)("Place the two left most terms in brackets")#

#" "(x^(color(magenta)(2)) +4x) -21#

#color(green)("Move the index of "color(magenta)(2)" to the outside of the brackets")#

#" "(x+4color(red)(x))^(color(magenta)(2))-21#

#color(green)("Remove the "color(red)(x)" from the "4x" inside the brackets")#

#" "(x+color(blue)(4))^2-21#

#color(green)("Apply "(1/2)xxcolor(blue)(4)=color(red)(2))#

#" "(x+color(red)(2))^2-21#

#color(green)("Add the constant "color(red)(k))#

#" "(x+2)^2+color(red)(k)-21#........................(2)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("Determine the value of "k)#

Set equation (1) = equation (2)

#" "x^2+4x-21=(x+2)^2+k-21#

#" "cancel(x^2)+cancel(4x)-cancel(21)=cancel(x^2)+cancel(4x)+4+k-cancel(21)#

#" "k=-4#

So for #(x+b)^2+k+c" "-> k=-b^2#

#" "(x+2)^2-25=0" "........................(2_a)#

So #color(blue)("Vertex"->(x,y) =((-1)xx2,-25) = (-2,-25))#

'~~~~~~~~~~~~~~~~~~~~~~~~~~

Tony B