How do you solve using the completing the square method #x^2 + 3x - 10 = 0#?

1 Answer
Apr 13, 2016

Answer:

#color(blue)(y_("intercept")=-10)#

#color(blue)("Vertex" -> (x,y)->(-3/2,-49/4))#

#color(blue)(x_("intercepts")=-5 or +2)#

Explanation:

Given:#" "x^2+3x-10=0#.......................(1)

#color(blue)("Determine "y_("intercept"))#

Read directly from equation (1)

#color(blue)(y_("intercept")=-10)#

#color(brown)("'~~~~~~~~~~~~~ Tip! ~~~~~~~~~~~~~~~~~~~~")#
#color(green)("As the equation is already in the form ")#

#color(green)(y=a(x^2+b/a x) + c)#

#color(green)("In this case "a=1)#

#color(green)(x_("vertex")=(-1/2)xxb/a = -3/2)#

#color(brown)("'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~")#

#color(blue)("Step 1")#
Write equation (1) as #(x^2+3x)-10=0#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 2")#

Add the adjustment constant #k#
#(x^2+3x)-10+k=0#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 3")#
Move the power from #x^2# to outside the bracket
#(x+3x)^2-10+k=0#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 3")#

Remove the #x" from "3x#
#(x+3)^2-10+k=0#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 4")#

Multiply the 3 inside the bracket by #1/2#
#(x+3/2)^2-10+k=0#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 4")#
If we multiply out the bracket we end up with an additional term to those in the original equation. That term is #(3/2)^2# derived from #(?+3/2)(?+3/2)=(3/2)^2#. This term must be removed which is achieved by making #k=-(3/2)^2#

So now we have

#color(brown)((x+3/2)^2-10+k=0)color(green)(->(x+3/2)^2-10-(3/2)^2=0)#

#color(blue)("Completing the square "->)color(magenta)( (x+3/2)^2-49/4=0)#......................(2)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#" "color(blue)("Vertex" -> (x,y)->(-3/2,-49/4))#

From equation (2)

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

Square rooting both sides

#x+3/2= +-sqrt(49)/sqrt(4)#

# color(blue)(x_("intercepts")=-3/2+-7/2 = #-5 or +2)#
Tony B