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My question is: Solve $(x-a)(x-b)(x-c)=0$ where $a,b,c$ belong to real numbers.

By observing, I found out that $x$ can be $a$ or $b$ or $c$.

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It's perfectly reasonable to say that $(x-a)(x-b)(x-c) = 0$ precisely when at least one of the terms is zero, because a product of three nonzero numbers is nonzero. Is this what you're looking for? By "method to solve", do you mean something like the quadratic formula? –  Ben Jun 4 '12 at 14:29

5 Answers 5

up vote 6 down vote accepted

In the real numbers, if a product $xy$ is zero (look in your textbook to find this, usually near the "factor theorem"), then one of the terms $x$ or $y$ is zero. So if $(x-a)(x-b)(x-c)=0,$ then either $x-a=0$, $x-b=0$ or $x-c=0$. Solve these and you will find $x$ is $a$, $b$ or $c$ respectively.

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Remember that the real numbers are an integral domain i.e. a commutative ring with no zero divisors. Hence, if we have a product $rs = 0$, this means that $r=0$ or $s=0$. Hence, if you have $$(x-a)(x-b)(x-c) = 0,$$ then either $(x-a)=0$ or $(x-b)=0$ or $(x-c)=0$. Adding $a$, $b$ and $c$ to the three equations respectively gives that $$x =a \text{ or }x =b \text{ or }x =c$$

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As the other answers mention, your solution is correct.

The reason for this is the "Zero product property", which applies because the real numbers are a domain (or integral domain).

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thanks for that yes i got it.Will you please tell me how to solve:|x-2|+|x-5|=3 –  meg_1997 Jun 4 '12 at 16:36
    
@meg_1997 You should ask that as a separate question. –  user17762 Jun 4 '12 at 16:36
    
You should ask that in a separate question. Or google "solving absolute value equalities". The general principle is to use the definition. –  The Chaz 2.0 Jun 4 '12 at 16:36
    
ya i tried but it says it does not meet our quality standards –  meg_1997 Jun 4 '12 at 16:37
    
What @Marvis |sivraM@ said... –  The Chaz 2.0 Jun 4 '12 at 16:38

If $$x_1*x_2*x_3=0,$$ then $$x_1=0 \quad\text{ or }\quad x_2=0 \quad\text{ or }\quad x_3=0.$$ In your case, $$x-a=0 \quad\text{ or }\quad x-b=0 \quad\text{ or }\quad x-c=0.$$ Thus $$x=a \quad\text{ or }\quad x=b \quad\text{ or }\quad x=c.$$

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1  
You know, it's not illegal to put more than one equals sign on a line :P –  Ben Millwood Jun 4 '12 at 15:20
    
I approved M Turgeon's edit to this answer. But the use of \mbox{} in the context is (IMO) an abomination, and I changed it to \text{}. See this....um....essay: en.wikipedia.org/wiki/… It is not only to Wikipedia that those remarks apply. –  Michael Hardy Jun 4 '12 at 16:26

My interpretation of your question is this: you've found that $x=a,b,c$ solve your equation, and you're curious about whether these are the only solutions to the equation. Suppose that $x$ is not one of $a,b,c$. Then all of the terms $(x-a), (x-b), (x-c)$ are not zero. If you multiply two real numbers that are not zero, then you will end up with something that is not zero. So $(x-a)(x-b)$ is not zero. Similarly, $(x-a)(x-b)(x-c)$ is not zero. This means that any $x$ that is not one of $a$, $b$, or $c$ will not solve your equation.

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