Why is it not D?
Sq.rt[(x-3)^2] = (3-x)
Squaring both sides, we get (x-3)^2 = (3-x)^2 and if you use any value the equation is satisfied right?
What am I missing here?
Sq.rt[(x-3)^2] = (3-x)
Squaring both sides, we get (x-3)^2 = (3-x)^2 and if you use any value the equation is satisfied right?
What am I missing here?
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