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I'm learning algebra (at age $66$) with home schooling materials. I am doing well and enjoying it. My question is this: The material says I can use "Completing the square" to solve any quadratic equation. I solved this equation $$6x^2+24x=0$$ by factoring (got answers $0$ and $-4$). But when I try and solve the same equation by completing the square I get a different answer ($\pm 12$). I must be doing something wrong but can't figure it out as I think I'm using the right steps to complete the square. Any help appreciated.

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  • $\begingroup$ Hint: $6(x^2 + 4 x + 4 - 4)\,$. $\endgroup$ – dxiv Jan 17 '17 at 20:35
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    $\begingroup$ Can you show us your steps to complete the square? $\endgroup$ – littleO Jan 17 '17 at 20:35
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    $\begingroup$ I took 1/2 of 24 (12) and squared it. Added that to both sides to get: 6X^2+24X+144 = 144. I then took 1/2 of 24 to get: (x+12)^2 = 144. Then took square root of (X+12)^2 so (x+ 12) equaled square root of 144. Then I used undoing to get x = square root of 144 - 12. So I ended up with X = two square roots of 144 (12 and - 12) and subtracted 12 from each to get answers 0 and -24. Sorry if I'm making this confusing. $\endgroup$ – James Ward Jan 17 '17 at 20:47
  • $\begingroup$ If you expand $(x+12)^2$ using FOIL, you can see it's not equal to $6x^2 + 24x + 144$. So, at that point, something has gone wrong. $\endgroup$ – littleO Jan 17 '17 at 20:52
  • $\begingroup$ I then took 1/2 of 24 That only works when the coefficient of $x^2$ is $1$. In your case you would take $\cfrac{1}{2} \ \cdot \cfrac{24}{6} = 2\,$ instead. $\endgroup$ – dxiv Jan 17 '17 at 20:53
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Your factoring solution is correct, and you can see it's correct by plugging in the solutions into the original formula. To complete the square, the first thing we want to do is divide both sides by the coefficient on the $x^2$ term, which is $6$ for this problem. Doing this gives $x^2 + 4x = 0$. Adding $4$ to both sides completes the square:

\begin{align} &x^2 + 4x = 0 \\ &x^2 + 4x + 4 = 4 \\ &(x+2)^2 = 4 \end{align}

To finish up, take a square root of both sides (and don't forget about the $\pm$ when doing this).


P.S. the quadratic formula $x = \displaystyle \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ that gives the solutions to an arbitrary $ax^2 + bx + c = 0$ can be derived with this completing the square method.

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  • $\begingroup$ Glad I could help! $\endgroup$ – Kaj Hansen Jan 18 '17 at 0:12
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Let us start with $6x^2+24x=6(x^2+4x)$. Skip $6$ for a moment to deal with $x^2+4x$. This is a part of a square $(x+2)^2$. If one square is $x^2$, then one term of the sqyare os a sum is $x$. Because double product is $4x,$ then the second summand should be $2$. Now we have $$(x+2)^2=x^2+4x+4.$$ To get $x^2+4x$ we need to subtract $4$: $$x^2+4x=(x+2)^2-4.$$ Finally $$6x^2+24x=6\bigl((x+2)^2-4\bigr)=6(x+2)^2-24.$$

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From the comments: Your problem is the $a$ value in $ax^2+bx+c$. When you're completing the square, the conventional method of approach is to divide both sides by $a$, to get$$ax^2+bx+c=0\implies x^2+\dfrac bax+\dfrac ca=0\tag1$$ This reduces the leading coefficent (was $a$) down to $1$, and you can continue as before. Hence,$$\begin{align*} & 6x^2+24x=0\\ & \implies x^2+4x=0\\ & \implies x^2+4x+\left(\dfrac 42\right)^2=\left(\dfrac 42\right)^2\\ & \implies (x+2)^2=4\\ & \ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\ldots\end{align*}$$ And continue as before.


Something worth noting:

This is exactly how the quadratic formula can be found. Start with the general quadratic $ax^2+bx+c=0$ and divide both sides by $a$. Then complete the square, and solve for $x$.

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