Solve using complete the square vs. quadratic formula

  • #1
paulb203
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24
Homework Statement
Solve for x using complete the square

x^2+6x=0
Relevant Equations
x=-b+/- sqrt b^2+4ac/(2a)
Using the complete the square method I got;

-3+root8 or -3-root8

But using the quadratic formula (to check my answer) I got;

-3+root 10 or -3-root 10

I've checked both answers several times but can't get to the bottom of it :)
 
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  • #2
paulb203 said:
Homework Statement: Solve for x using complete the square

x^2+6x=0
Relevant Equations: x=-b+/- sqrt b^2+4ac/(2a)

Using the complete the square method I got;

-3+root8 or -3-root8

But using the quadratic formula (to check my answer) I got;

-3+root 10 or -3-root 10

I've checked both answers several times but can't get to the bottom of it :)
You can substitute your answers in the original equation to check.
You will find that all of them are wrong.
 
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  • #3
paulb203 said:
Homework Statement: Solve for x using complete the square

x^2+6x=0
Relevant Equations: x=-b+/- sqrt b^2+4ac/(2a)

Using the complete the square method I got;

-3+root8 or -3-root8

But using the quadratic formula (to check my answer) I got;

-3+root 10 or -3-root 10

I've checked both answers several times but can't get to the bottom of it :)
Your relevant equation misses brackets:$$x = {-b\pm\sqrt{b^2-4ac}\over 2a}$$

Both your answers don't tell us what you did wrong, only that it came out wrong.

The exercise wants you to complete the square -- so show us how you do that.

(Of course nothing prohibits you to solve in a different way so you have the correct answers for checking purposes :wink:)

##\ ##
 
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  • #4
"Inspection" is an underestimated mathematical technique!
 
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  • #5
paulb203 said:
Using the complete the square method I got;

-3+root8 or -3-root8

But using the quadratic formula (to check my answer) I got;

-3+root 10 or -3-root 10
I have no idea how you came up with either "root8" or "root10".

As a side note, there's a link to our LaTeX tutorial at the lower left corner of the input pane. Instead of writing, say, "root 10" you can format it very nicely as ##\sqrt{10}##.
Before it's rendered by a browser, it looks like this: ##\sqrt{10}##.

Your original equation looks like this: ##x^2 + 6x = 0##.
What I wrote was this: ##x^2 + 6x = 0##.
BvU said:
Of course nothing prohibits you to solve in a different way so you have the correct answers for checking purposes :wink:
And the given equation can be solved by a very simple "different way."
 
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  • #6
paulb203 said:
Homework Statement: Solve for x using complete the square

x^2+6x=0
Relevant Equations: x=-b+/- sqrt b^2+4ac/(2a)
You should use parenthesis to make this equation clear. They don't cost you anything and they might help.
paulb203 said:
Using the complete the square method I got;

-3+root8 or -3-root8

But using the quadratic formula (to check my answer) I got;

-3+root 10 or -3-root 10

I've checked both answers several times but can't get to the bottom of it :)
You don't show any of your work to get to those "answers", which are all wrong. So we can't help much.
 
  • #7
Completing the square means rewriting ##x^2+ 6x## by adding and subtracting a new term, so the resulting expression is of the form ##(ax+b)^2+c##. Your terms start looking like ##x^2+2x## and look like the previous when rewritten with the additional terms.
 
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  • #8
WWGD said:
Completing the square means rewriting ##x^2+ 6x## by adding and subtracting a new term, so the resulting expression is of the form ##(ax+b)^2+c##.
Or by adding the same amount to both sides of the given equation so as to give a perfect square trinomial on the left side of the equation.
##x^2 + 6x + ? = 0 + ?##
 
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  • #9
Thanks, guys.
I don't know how I got those answers! I can't find my original working
I redid it and got x=0, or, x=-6
 
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  • #10
Here's how it goes with completing the square:
##x^2 + 6x = 0##
##\Rightarrow x^2 + 6x + 9 = 9##
##\Rightarrow (x + 3)^2 = 9##
##\Rightarrow x + 3 = \pm \sqrt 9 = \pm 3##
##\Rightarrow x = -3 \pm 3##
##\Rightarrow x = 0## or ##x = -6##

A third method other than the two methods mentioned in the problem statement is solving by factorization.
##x^2 + 6x = 0##
##\Rightarrow x(x + 6) = 0##
##\Rightarrow x = 0## or ##x = -6##
 
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  • #11
@paulb203, I hope that you realize that the quadratic formula is easily derived by completing the square. So the two are actually equivalent.

Suppose ##ax^2+bx+c = 0##, where ##a \ne 0##. Completing the square in a few steps:
##x^2 + \frac{b}{a} x +\frac {c}{a} = 0##
##x^2 +\frac{b}{a} x = -\frac{c}{a}##
## x^2 + \frac{b}{a} x + (\frac {b}{2a})^2 = -\frac{c}{a} + (\frac {b}{2a})^2##
##(x + \frac {b}{2a})^2 = -\frac{c}{a} + (\frac {b}{2a})^2##
##x + \frac {b}{2a} = \pm \sqrt {-\frac{c}{a} + (\frac {b}{2a})^2}##
##x = -\frac {b}{2a} \pm \sqrt {-\frac{c}{a} + (\frac {b}{2a})^2}##
##x = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}##
 
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  • #12
Thanks, guys.
 

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