How do you solve using the completing the square method #r^2 + 14r = -13#?

3 Answers
Apr 15, 2017

#r^2 + 14r = -13#

#implies r^2 + 14r + 13 = 0#

We could have a go at factoring that, but completing the square is the next step down the list and is what you are asking for.

It is the Quadratic Equation without having to remember a formula!

So:

# r^2 + 14r + 13#

#= r^2 + (color(red)(2 xx) 7)r + 13 #

[Remember that: #(a + b)^2 = a^2 + color(red)(2) ab + b^2#]

#= (r + 7)^2 - 7^2 + 13 #

#= (r + 7)^2 - 36 #

Now:

#(r + 7)^2 - 36 color(red)(= 0)#

#implies (r + 7)^2 = 36 #

#implies r + 7 = sqrt36 #

#implies r =- 7 pm 6 = -1, -13#

We can check that:

#(r+1)(r+13) = r^2 + 13r + r + 13 color(blue)(= r^2 + 14r + 13)#

Apr 15, 2017

Answer:

See the method outlined below:

Explanation:

First, write it in general form:

#r^2+14r+13=0#

Add bracket around the terms that contain #r#, with the constant term outside the bracket (and factor out the #a# coefficient if there is one):

#(r^2+14r)+13=0#

Now, take half of the value of #b# (the coefficient of the term in #r#) and square it. Add this value and subtract it as well, within the bracket.

#(r^2+14r+49 - 49)+13=0#

Move the subtracted #(b/2)^2# term outside the bracket. (This can require some care if there was an #a# value (a coefficient of the #r^2# term greater than 1.)

#(r^2+14r+49) - 49+13=0#

By design, the part on the brackets is a perfect square:

#(r+7)^2 -36 = 0#

It is now in standard form, so you are done.

The vertex is at (-7, -36).

Apr 15, 2017

Answer:

#r={-1,-13}#

Explanation:

#r^2+14r=-13#

#"Let us rearrange the equation."#

#r^2+14r+13=0#

#"let us add+36 -36"#

#r^2+14rcolor(red)(+36-36)+13=0#

#r^2+14r+36+13color(red)(-36)=0#

#color(green)(r^2+14r+49)-color(red)(36)=0#

#color(green)(r^2+14r+49)=(r+7)^2#

#(r+7)^2-36=0#

#(r+7)^2-6^2=0#

#"let us remember " a^2-b^2=(a-b)(a+b)#

#(r+7-6)(r+7+6)=0#

#(r+1)(r+13)=0#

#"Either (r+1) or (r+13) is equal to zero."#

#r+1=0 rArr r=-1#

#r+13=0rArr r=-13#