How do you solve using the completing the square method #x ^2 + 13 x + 42 = 0#?

1 Answer
Mar 20, 2016

Answer:

For a very detailed example of method see
http://socratic.org/s/asVauX76

#"Vertex" ->(x,y)=(-13/2 -1/4) = (-6 1/2,-1/4)#
I will let you solve the other bits!

Explanation:

Given:#" "x^2+13x+42=0#

The process applied introduces an error that is removed by adding the correction factor of #k#

Write as:#" "(x^2+13x)+42=0#

Add the correction factor of #k#

#" "(x^2+13x)+42+k=0#

Take the power to outside the brackets

#" "(x+13x)^2+42+k=0#

Halve the coefficient of #13x#

#" "(x+13/2 x)^2+42+k=0#

Remove the #x# from #13/2 x#

#" "(x+13/2)^2+42+k=0#

The error comes from #(+13/2)^2 =+ 169/4#

So #k=-169/4# giving:

#" "(x+13/2)^2+42-169/4=0#

#" "(x+13/2)^2 -1/4=0#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#"Vertex" ->(x,y)=(-13/2 -1/4) = (-6 1/2,-1/4)#

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

Find x-intercepts by substituting 0 for y and solve in the normal way
y-intercept is the constant in the original equation = +42

I will let you solve for the x-intercepts

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