Confusion about solving quadratics $x^2-8x=5$

Question

I got into a little confusion, hopefully someone can clear it up for me. So I'm solving $x^2-8x=5$ right, and here are my steps:

$$1: x^2-8x=5$$ $$2: x^2-8x+16=21$$ $$3: (x-4)^2=21$$ $$4: x-4=\pm\sqrt{21}$$ $$x=\pm\sqrt{21} +4$$

So I have a question, why is it plus or minus? I thought the square root is usually positive?

Answer 1

Every positive number $n$ has two square roots: $\sqrt{n}$ and $-\sqrt{n}$. Both $4 + \sqrt{21}$ and $4 - \sqrt{21}$ are solutions to this equation. You can try plugging them in with your calculator to verify.

What you're thinking of here is that $\sqrt{n}$ refers only to the positive square root. If you want the other square root, you must specify $-\sqrt{n}$. This way, we never need to be confused when we see $\sqrt{n}$, even though $n$ has two roots.

As for your other question, it is not true that $\sqrt{x} + y = \sqrt{x + y}$, which is what you're observing here.

Answer 2

There can be two square roots of every positive number: one positive and the other negative. For example, $ \sqrt 25 = ± 5$ which is true as $ 5^2 = 25 $ and also, $ (-5)^2 = 25 $.

Yes, the square root is usually positive ( $ \sqrt x $ denotes the positive root of $x$ only), but square roots can be negative as well. The above example proved it already. Also, in the fourth step you should write,

$$ (x-4) = ± \sqrt 21 $$

Answer 3

First, why is it plus or minus?

Because the equation $\,x^2=b\,$ has two solutions $\,x = \,\pm \sqrt{b}\,$ when $\,b \gt 0\,$.

I thought the square root is usually positive?

That's correct. The real square root function $\,\sqrt{\,\cdot\,}\,$ returns the positive solution, by definition.

Second, and this really confuses me, we can see that in step 4 I removed the square from the factored side, what allows me to take the square root of the other side?

Suppose $\,a^2 = 21\,$, then that's equivalent to $\,a^2 - 21=0\,$, which using the algebraic identity $\,a^2-b^2=(a-b)(a+b)\,$ with $\,b=\sqrt{21}\,$ is in turn equivalent to $\,(a - \sqrt{21})(a+\sqrt{21})=0\,$. Therefore one of the factors must be $\,0\,$, so either $\,a=\sqrt{21}\,$ or $\,a=-\sqrt{21}\,$, which is usually written as $\,a = \pm \sqrt{21}\,$. Your question is about the case where $\,a = x-4\,$.

Answer 4

We have the equation $$x^2 - 8x = 5$$ Add $16$ to both sides so we can factor the LHS in a nice way, so \begin{align} x^2 - 8x + 16 &= 5 + 16 \\ (x-4)^2 &= 21 \end{align} Now take the square root on both sides (we can do this because both sides of the equation are positive terms)

$$ \sqrt{(x-4)^2} = \sqrt{21}$$

Notice that $x$ could be a positive or negative number and still make the equation true, we have to consider such cases, so

\begin{align} x-4 &= \pm \sqrt{21} \\ x &= 4 \pm \sqrt{21} \end{align}

will be the solutions to the quadratic equation.