# (Highschool Pre-calculus) Solving quadratic via completing the square

I'm trying to solve the following equation by completing the square:

$x^2 - 6x = 16$

The correct answer is -6,1. This is my attempt:

$x^2 - 6x = 16$

$(x - 3)^2 = 16$

$(x - 3)^2 = 25$

$\sqrt(x -3)^2 = \sqrt(25)$

$x - 3 = \pm5$

$x =\pm5 - 3$

$x = -8,2$

I did everything according to what I know,but my answer was obviously wrong. Any help is appreciated. Thanks

• You switched the sign. Should be $8,-2$ Jan 17 '14 at 12:53
• @in_wolfram_we_trust: I have rolled back your edit. Please do not edit questions to fix things the asker did wrong; this will obscure information that may be necessary to write a helpful answer to them. Jan 17 '14 at 13:21

If the solutions were supposed to be $-6$ and $1$, the problem would've been $$(x + 6)(x - 1) = 0\\ x^2 + 6x -x - 6 = 0\\ x^2 + 5x = 6$$ Either you've copied the wrong problem, the wrong solution, or there is a mistake in your solution collection.

I am just going to do the thing first and we will talk about what you did after.

Note that $x^2-6x\sim (x-3)^2$. In fact

\begin{align}(x-3)^2=x^2-6x+9&=(x^2-6x)+9\\\Rightarrow x^2-6x&=(x-3)^2-9.\end{align}

Therefore if $$x^2-6x=16$$ then \begin{align}(x-3)^2-9&=16 \\\Rightarrow (x-3)^2&=25 \\ \Rightarrow x-3&=\pm5 \\ \Rightarrow x&=8\text{ or }-2. \end{align}

The problems with what you did are as follows:

1. If $x^2-6x=16$ it does not follow that $(x-3)^2=16$.
2. You don't need to take the square roots of both sides necessarily. If $x^2=a$ then $x=\pm \sqrt{a}$ is a perfectly good implication.
3. At the end you have two equations... $x-3=5$ and $x-3=-5$. To get rid of the three, remember you want $x=\dots$ add three to both sides, e.g. \begin{align} x-3&=5 \\\Rightarrow x-3+3&=5+3 \\\Rightarrow x+0&=8 \\ \Rightarrow x&=8. \end{align}

You can do this because if two quantities are equal, and you add the same thing to both of them, then they are still equal.

• Hmm, okay I see. Just to confirm, the answer is indeed 8,-2 , yes ? My answer sheet says -6, 1. Thank you for your help, and everyone else too. Also regarding point #2, you said we don't need to take the square roots of both sides but from my understanding, we need to keep the equation balanced so we must apply the same operation to both sides, ? Jan 17 '14 at 13:16
• @Sophia Your understanding about point #2 is correct, but what Jp means is, you don't need to write it down as a separate step. Jan 17 '14 at 13:40
• Ok thanks @Arthur for clearing that up. Jan 17 '14 at 13:42

$x^2 - 6x$ isn't the same as $(x-3)^2$, so your equation $(x-3)^2 = 16$ is wrong.

$(x-3)^2$ is actually $x^2 - 6x + 9$, so you should write $$x^2 - 6x + 9 = 25$$ and then $$(x-3)^2 = 25$$. So your third equation is correct, even though your second wasn't.

I don't agree with your equation $\sqrt{(x-3)^2} = \sqrt{25}$. Technically it is correct, but you should know that in general $\sqrt{A^2}$ is not always $A$. In general, $\sqrt{A^2} = |A|$. So taking the square roots of both sides is not a good way to explain this. Instead, write $x-3 = \pm \sqrt{25}$. (It is a fact that if $z^2 = a$ and $a \geq 0$, then $z = \pm \sqrt{a}$.)

Your main mistake, and the only one that leads to an error in the result, is where you add 3 to the left side of your equation, but subtract 3 from the right side, and obtain $x = \pm 5 - 3$. You should instead add 3 to both sides. You need to do the same thing to both sides.

• She's not rewriting $x^2-6x$ to $(x-3)^2$, but to $(x-3)^2-9$, which seems to be quite correct. Jan 17 '14 at 13:14
• There have been edits to her answer. Jan 17 '14 at 13:18
• x @user121926: Ah, sneaky! I've rolled them back. Jan 17 '14 at 13:22

no ,no,what do you need is following

we have

$x^2-6*x=16$

$x^2-6*x+9=16+9$

$(x-3)^2=25$

now $x-3=5$ or $x=8$

and $x-3=-5$ or $x=-2$

there is no mistake,why is answer $-6$?

$(-6)^2-6*(-6)=36+36=72$