Solve using complete the square vs. quadratic formula

[paulb203]

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 :)

 

[Hill]

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.

 

[BvU]

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 )

$\$

 

[PeroK]

"Inspection" is an underestimated mathematical technique!

 

[Mark44]

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: $x^2 + 6x = 0$.

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

And the given equation can be solved by a very simple "different way."

 

[FactChecker]

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.

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.

 

[WWGD]

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.

 

[Mark44]

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.
$x^2 + 6x + ? = 0 + ?$

 

[paulb203]

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

 

[Mark44]

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 $x = -6$

 

[FactChecker]

@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:
$x^2 + \frac{b}{a} x +\frac {c}{a} = 0$
$x^2 +\frac{b}{a} x = -\frac{c}{a}$
$x^2 + \frac{b}{a} x + (\frac {b}{2a})^2 = -\frac{c}{a} + (\frac {b}{2a})^2$
$(x + \frac {b}{2a})^2 = -\frac{c}{a} + (\frac {b}{2a})^2$
$x + \frac {b}{2a} = \pm \sqrt {-\frac{c}{a} + (\frac {b}{2a})^2}$
$x = -\frac {b}{2a} \pm \sqrt {-\frac{c}{a} + (\frac {b}{2a})^2}$
$x = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}$

 

[paulb203]

Thanks, guys.