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Quadratic equation

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Quadratic equation

by Mustang » Sat Apr 18, 2009 9:53 am
How do we solve this:

Sq root[(x-3)^2] = (3-x)

just by looking we can figure out that the two roots will be x=0 and x=3

but how can we get both the roots by solving it. I was able to get x=3 but wasn't able to get x=0.
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by DanaJ » Sun Apr 19, 2009 1:22 am
One way of solving this equation would be to remember that:

sqrt(x^2) = |x| or that the square root of x^2 will always be the absolute value of x.

Since you have that sqrt[9x - 3)^2] = 3 - x, this means that |x - 3| = 3 - x or that |x - 3| = -(x - 3). This only happens if x - 3 is negative or equal to zero (remember, one of the properties of absolute value is that |a| when a is negative will be -a). This is why you get:

x - 3 <= 0
x <= 3.

Your solution will be (-infinite, 3] - test any number within this interval for proof.

I must say you had me fooled with that x = 0 solution. While it's true that the solutions you provided ARE valid, they're not by a long shot the only ones... I struggled for a few minutes, trying to find a way to prove that x = 0 is a solution, and then it hit me... Hope you find my explanation useful.
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