Solve using complete the square vs. quadratic formula

In summary, solving quadratic equations can be approached using two main methods: completing the square and the quadratic formula. Completing the square involves rearranging the equation into a perfect square trinomial, making it easier to solve for the variable. On the other hand, the quadratic formula provides a direct solution for any quadratic equation in the standard form \(ax^2 + bx + c = 0\) using the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Each method has its advantages, with completing the square offering deeper insight into the properties of the equation, while the quadratic formula is often quicker for finding roots.
  • #1
paulb203
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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.
 

FAQ: Solve using complete the square vs. quadratic formula

What is the difference between completing the square and the quadratic formula?

Completing the square is a method of solving quadratic equations by transforming the equation into a perfect square trinomial, while the quadratic formula provides a direct solution using the coefficients of the quadratic equation. Both methods ultimately find the roots of the equation but use different approaches.

When should I use completing the square instead of the quadratic formula?

Completing the square is often used when you need to derive the quadratic formula, solve quadratic equations that are not easily factorable, or when you need to convert a quadratic equation into vertex form. The quadratic formula is generally quicker and more straightforward for solving standard quadratic equations.

Can completing the square be used for all quadratic equations?

Yes, completing the square can be used for all quadratic equations. However, it may be more cumbersome for some equations compared to using the quadratic formula. The quadratic formula is a more universally applicable and quicker method for solving any quadratic equation.

How do you complete the square for a quadratic equation?

To complete the square for a quadratic equation of the form ax^2 + bx + c = 0, follow these steps: (1) Divide all terms by 'a' if 'a' is not 1. (2) Move the constant term 'c' to the other side of the equation. (3) Add and subtract (b/2a)^2 inside the equation. (4) Rewrite the left side as a perfect square trinomial. (5) Solve for 'x' by taking the square root of both sides and then isolating 'x'.

What are the advantages of using the quadratic formula over completing the square?

The quadratic formula is advantageous because it provides a straightforward and quick solution to any quadratic equation without the need for additional steps like transforming the equation. It also works uniformly for all types of quadratic equations, making it a more versatile and less error-prone method compared to completing the square.

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