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As I was doing a SAT question when I came across this question:

$\sqrt {x-a} = x-4$        If $a=2$,what is the solution set of the equation?

Options

  • {$3,6$}
  • {$2$}
  • {$3$}
  • {$6$}     Correct Answer

I evaluated the equation and got $0=(x-3)(x-6)$
If you put those number in the equation, you should get:
For 3:
$\sqrt {3-2} = 3-4$
Since $\sqrt {1} = ±1$  
$±1 = -1$


For 6:
$\sqrt {6-2} = 6-4$
Since $\sqrt {4} = ±2$  
$±2 = 2$

For the answer, they(SAT) evaluated $\sqrt {1}$ as $\sqrt {1} = -1$ and $\sqrt {4}$ as $\sqrt {4} = 2$
Why is it that $\sqrt {1}$ is equal to $-1$ and not $1$ and why $\sqrt {4}$ is equal to $2$ and not $-2$ Why isn't the solution set {$3,6$} a correct answer?

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    $\begingroup$ This was an equation to solve in real numbers, so the symbol $\sqrt x $ is meant to be the *positive* square root. No $\pm $ - only $+$. Also note your error in case 3: $3-4=-1$, not $1$. $\endgroup$
    – user491874
    Commented Dec 1, 2017 at 0:36
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    $\begingroup$ 3 is not a solution precisely because $\sqrt 1\ne-1$. I think "they" must have written $\sqrt 1=-1$ as an obvious contradiction that you reach if you assume 3 is a solution, therefore 3 is not a solution. $\endgroup$
    – user491874
    Commented Dec 1, 2017 at 0:45

4 Answers 4

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By definition $\sqrt{}$ is always the non-negative root. Every positive number has exactly two square roots equal in magnitude, on positive and one negative.

$\sqrt{25} =5$ and $\sqrt {25} \ne -5$. But both $5$ and $-5$ are solutions to $x^2 = 25$.

To solve an equation $x^2 = k$ there will be two answers. One is $\sqrt k$ and $\sqrt k > 0$ and the other is $-\sqrt{k}$ and $-\sqrt {k} < 0$

So if you try to solve an equation by "squaring both sides", you will be changing the equation to allow for two different square roots that were not part of the original problem. This is called superfluous solutions.

So to solve

$\sqrt {x-2} = x- 4$

Is not just to solve $x-2 = (x-4)^2$ but is to ALSO solve $x - 4 \ge 0$.

So you did $\sqrt{x-2}^2 = (x-4)^2$. But that adds in the negative solution as well.

$x^2 - 8x + 16 = x-2$ and $x^2 - 9x + 18 = (x-6)(x-3)$ so both of those solve $x-2 = (x-4)^2$ but only one of them solve $\sqrt{x-2} = x-4$. (Because we must have $x-4 \ge 0$.)

$\sqrt{6-2} \overset?= 6-4$

$\sqrt{4} \overset?= 2$

$2 \overset \checkmark = 2$ check. $6$ is an answer and $6-4 > 0$.

$\sqrt{3-2} \overset?= 3-4$

$\sqrt{1} \overset?= -1$

$1 \ne -1$. No! $\sqrt{1} = 1$. $\sqrt{1} \ne -1$. And $3-2 < 0$.

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  • $\begingroup$ Can u explain the ?? ? $\endgroup$ Commented Dec 1, 2017 at 0:59
  • $\begingroup$ I meant "do they equal" $\endgroup$
    – fleablood
    Commented Dec 1, 2017 at 2:10
  • $\begingroup$ Ah I see. That makes sense $\endgroup$ Commented Dec 1, 2017 at 2:10
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    $\begingroup$ I hope writing $\overset?=$ was okay for you. I think the $=?=$ looks strange. I have no idea what $===$ and $=\not==$ stands for. Nice answer otherwise. $\endgroup$
    – M. Winter
    Commented Dec 1, 2017 at 13:10
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    $\begingroup$ MathJax Griping: so learning of \overset I figured a checkmark over the = sign would be good. So I did \overset \check = ... and got a "missing argument for check" error. google search and I find that \checkmark is $\checkmark$. So now I wonder, what is check and what the argments for check are suppose to be. Anyway $\overset \checkmark =$ is kind of cool looking. $\endgroup$
    – fleablood
    Commented Dec 1, 2017 at 16:38
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The standard way to solve equations with square roots is to use this rule: $$ \sqrt A=B\iff (A=B^2\quad\textbf{and}\quad B\ge 0). $$ So here you obtain $$\sqrt{x-2}=x-4\iff x-2=x^2-8x+16\;\text{and}\; x\ge 4\iff(x-3)(x-6)=0\;\text{and}\; x\ge 4,$$ which shows there's only one root: $6$.

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  • $\begingroup$ Why can't B be less than zero? $\endgroup$ Commented Dec 1, 2017 at 0:50
  • $\begingroup$ Also, where did u get x ≥ 4? $\endgroup$ Commented Dec 1, 2017 at 0:53
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    $\begingroup$ Because, by convention, the symbol $\sqrt{\text{positive number}}$ denotes, of the two square roots of a positive number, the positive root. $x\ge4$ comes from $B=x-4 \ge 0$. $\endgroup$
    – Bernard
    Commented Dec 1, 2017 at 0:53
  • $\begingroup$ The brave anonymous down-voter struck again! $\endgroup$
    – Bernard
    Commented Dec 1, 2017 at 17:21
  • $\begingroup$ Really, what's going on is $$ \sqrt A=B => A=B^2\quad. $$ So once we solve the second equation, we test them in the first equation to see if they're valid. $\endgroup$ Commented Dec 1, 2017 at 19:30
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The square root sign over a real number always denotes just the nonnegative square root. There's no $\pm$. If you square an equation with a square root in it you run the risk of allowing an extraneous solution.

In this example, $\sqrt{3-2} = 1 \ne 3-4$ .

There are many questions on this site with this answer, It was just easier for me to write it than to find them.

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You factored it correctly, but your mistake is considering both roots of $x$. When we say $\sqrt{x}$, we refer to the principal root of $x$; that is, the $positive$ square root of $x$ (issue in red, corrections in blue):


For 3:
$\sqrt {3-2} = 3-4$
$\sqrt {1} = \color{red}{±}1$
$\color{blue}{\sqrt {1} = -1}$
$\color{blue}{1 = -1 \, \, \, \, \, \, \, \text{false; not in solution set}}$


For 6:
$\sqrt {6-2} = 6-4$
$\sqrt {4} = \color{red}{±}2$
$\color{blue}{\sqrt {4} = 2}$
$\color{blue}{2 = 2 \, \, \, \, \, \, \, \text{true; in solution set}}$

The extraneous solution $x=3$ was brought about when you squared the radical, because squaring removes the restriction that $\sqrt{u} \geq 0$.

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    $\begingroup$ To be more precise, squaring the equation removes the condition $x-4\geq0$, or $x\geq4$ $\endgroup$
    – Nick2253
    Commented Dec 1, 2017 at 5:08

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