How do you solve #x^2-5x-6=0# by completing the square?

2 Answers
Sep 7, 2016

Answer:

#x=-1" and "x=6#

Please do not change this solution. It is a reference solution.

Explanation:

#color(blue)("End objective")#

Given the standard form #y=ax^2+bx+c#

We end up with form #y=a(x+ b/(2a))^2+c+k#

The problem is that changing the original equation to this form introduces an error. So #k# is a necessary correction. The value of which is determined by:

#a(b/(2a))^2+k=0#

#color(magenta)("For this question "a=1)#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Completing the square")#

Given:#" "y=0=x^2-5x-6#.......................Equation(1)

#color(brown)("Step 1")#

Write as: #(x^2-5x)-6+k larr" At this stage "k" has no value"#

'...................................................................................................
#color(brown)("Step 2")#
Move the power from #x^2# to outside the brackets

#(x-5x )^(color(magenta)(2))-6+k #

#color(purple)(" Now "k" starts to take on values that change at each step")#

'...................................................................................................
#color(brown)("Step 3")#
Discard the #x# from #5x# inside the brackets

#(x-5)^2-6+k#

'...................................................................................................
#color(brown)("Step 4")#

Halve the 5 inside the brackets

#(x-5/2)^2-6+k#

'...................................................................................................
#color(brown)("Step 5")#

Determine the value of #k#
#(xcolor(magenta)(-5/2))^(color(magenta)(2))-6+color(magenta)(k)#......................Equation(2)

Let #(color(magenta)(-5/2))^(color(magenta)(2))+ color(magenta)(k)=0" "rarr" Note that this is the bit:"#
#color(white)("222222222222222222222222") "a(color(magenta)(b/(2a)))^(color(magenta)(2))+color(magenta)(k)=0#

#color(white)("ddddddddddddddddddddd")#where in this case #a=1#

#=>25/4+k=0" " =>" " k= -25/4#

So by substitution equation(2) becomes:

#y=0=(x-5/2)^2-6-25/4#

#color(blue)(y=0=(x-5/2)^2-49/4)color(white)(.)......Equation(2_a)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("Step 6 - Determine the vertex")#

Using #Eqn(2_a)#
Vertex#->(x,y)=[(-1)xx(-5/2)color(white)("ddd"),color(white)("d")-49/4]#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Solving for "y=0)#

Add #49/4# to both sides

#(x-5/2)^2=49/4#

Square root both sides

#x-5/2=+-sqrt(49/4) = +-7/2#

Add #5/2# to both sides

#x=5/2+-7/2#

#x=-1" and "x=6#

Tony B

Dec 5, 2017

Answer:

As Tony B. states the answer is #x=6# or #x=-1#. The explanation here uses a different method to complete the square.

Explanation:

In this method a different approach is used to complete the square. The idea is to show more clearly how the coefficient #5# gets "halved".

Start with the expression

#x^2-5x=x(x-5)#

These expression is the same as the first two terms of the quadratic expression originally given. We render this as a difference of squares:

#x(x-5)=u^2-v^2=(u+v)(u-v)#

And then we match the two sets of factors:

#x=u+v#

#x-5=u-v#

Now add these two equations together and note that in eliminating #v# we make a coefficient of #2# on #u#. This is how we end up "halving the #5#:

#2x-5=2u# so #u=(x-5/2)#

Then take the difference of the original two equations to get #v# which is just a constant, since #u# and #x# cancel at the same time:

#5=2v# so #v=5/2#; again half of #5# appears.

This works because the original quadratic expression had a coefficient of #1# on #x^2#. If the original quadratic expression has a different coefficient, like #2# in #2x^2-10x-12=0#, we would have divided by that coefficient on #x^2# at the beginnjng to get (in this case) #x^2-5x-6=0# with a coefficient of #1# on #x^2#, and then we could proceed as above.

Then we have:

#x^2-5x=u^2-v^2=(x+5/2)^2-({25}/4)#

And the original equation becomes:

#x^2-5x-6=0#

#(x-5/2)^2-({25}/4)-6=0#

#(x-5/2)^2=({49}/4)#

So we take both possible square roots noting that #{49}/4# happens to be a perfect square:

#(x-5/2)=+(7/2)# so #x=6# or

#(x-5/2)=-(7/2)# so #x=-1#

If we did not have a perfect square the final answer would contain square roots.