Are these steps for finding the solutions of $\sqrt{x + 3} = 3 − x$ correct?

  1. $\sqrt{x + 3} = 3 − x$ is given;
  2. $x + 3 = x^2 − 6x + 9$, obtained by squaring both sides of (1);
  3. $0=x^2 −7x+6$, obtained by subtracting $x+3$ from both sides of (2);
  4. $0 = (x − 1)(x − 6)$, obtained by factoring the right-hand side of (3);
  5. $x = 1$ or $x = 6$, which follows from (4) because $ab = 0$ implies that $a=0$ or $b=0$.
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    $\begingroup$ I don't understand ab = 0 implies that a=0 or b = 0 can someone explain $\endgroup$ – Bogle Feb 27 '17 at 23:01
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    $\begingroup$ I formatted your question; could you please read through it and check that it says what you want it to say? $\endgroup$ – Patrick Stevens Feb 27 '17 at 23:07

Yes, all the steps are correct. You only need to check those value of $x$ if it satisfies the original equation or not.

Assume that $ab=0$. We want to show that either $a=0$ or $b=0$. If $a=0$ then we are done. Suppose that $a\neq 0$. Then $a^{-1}\in\Bbb R$. Thus, using field properties of $\Bbb R$ we get $$b=1b=(a^{-1}a)b=a^{-1}(ab)=a^{-1}0=0.$$

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    $\begingroup$ To the down voter, whats the basis of down voting my answer? Please let me know. $\endgroup$ – Juniven Feb 28 '17 at 0:58


x = 1 is a correct answer to the problem. If you substitute it into the original question, you get 2 = 2, correct.

x = 6 has a little problem. When you substitute it into the original question, you get 3 = -3. Oops.

When you square an equation, you lose the sign information. Both positive and negative roots will come out as answers. But in the original problem we assume positive roots.

Here 6 is called an extraneous root. Your mathematics is right but you missed the vital final step -- you MUST check answers in the original equation.

As far as ab = 0 implies either a = 0 or b = 0. this is a basic property of real numbers which you know well. Consider your multiplication tables and also decimals or fractions. What is the only way you get zero as a product?

Finally your work is good and organized. The editing done above is also extremely important. You must do one step per line, only, and put each line directly below the one you derived it from. That way people can read your work. If it as mashed together nobody can make sense out of it.


Those steps are correct, but there's one final step required:

  1. When $x=1$, the left-hand side is $2$ and the right-hand side is $2$; when $x=6$, the left-hand side is $3$ and the right-hand side is $-3$. So only the first "solution" is a solution. (Er. When I did this the first time, I wrongly thought that the RHS in the $x=6$ case was $3$.)

To explain step 5: we want to find $x$, given that $(x-1)(x-6) = 0$. Let $a=x-1$ and $b=x-6$; then since $ab=0$, we have $a=0$ or $b=0$. So $x-1=0$ or $x-6=0$; so $x=1$ or $x=6$.

Why is it the case that if $ab=0$ then $a=0$ or $b=0$? This is precisely the statement that the product of nonzero quantities is nonzero (by taking the contrapositive), which you might find more intuitive; but I see there is another answer which gives algebraic manipulations to prove it.


It's only valid to square the equation when both sides are positive.

In your second line you need to have a condition $3-x\geq 0 $ which simplifies to $x\leq 3$.Also you must have your square root defined i.e $x+3\geq 0$ or $x\geq -3$.

Now that we've done that we can see that $x=6$ is not a solution to the equation,this can be easily verified by plugging $x=6$ or differently $\sqrt{6+3}\neq 3-6$


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