anudoneddbv

2022-07-17

Discrete Math Proofs
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. $0=\left(x-1\right)\left(x-6\right)$, 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 $b=0$.

Expert

Step 1
Those steps are correct, but there's one final step required:
6. 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 $\left(x-1\right)\left(x-6\right)=0$. 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 $x=6$.
Step 2
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.

Damien Horton

Expert

Step 1
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 $b=0$.
Step 2
If $a=0$ then we are done. Suppose that $a\ne 0$. Then ${a}^{-1}\in \mathbb{R}$. Thus, using field properties of R we get $b=1b=\left({a}^{-1}a\right)b={a}^{-1}\left(ab\right)={a}^{-1}0=0.$

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