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# help

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Solve $$|x-3| + \sqrt{x^{2}-9} = 0$$

Jun 28, 2020

#1
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Hi there! I have figured out the answer but dont really know how to put it into words...

lets try though!

So we know that $$|x-3|$$ will have to be positive because of the absolute value.

and if we expand out the radical $$\sqrt{x^2-9}$$ we get$$\sqrt{(x-3)(x+3)}$$

Now that we have this we can subtract $$|x-3|$$ from the left side. This means that our new equation is

$$\sqrt{(x-3)(x+3)}=-|x-3|$$

note: because of the absloute value we can get rid of the negitive

Because there is a radical we can square both sides to get

$$(x-3)(x+3)=|x-3|^2$$

We have to disbtrube the left side so

$$x^2+3^2=(x-3)^2$$

simpify $$3^2=9$$ you can solve from there!

also this might not be the best way to solve a question like this because it is my personal way aswell the way I learned it.

Hope this helps

~Wolf :D

Jun 28, 2020
#2
+27660
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Hmmmm..... I am not sure this is possible....

sqr (x^2-9)  will always be positive     which would mean  |x-3| must be NEGATIVE for the terms to ADD to 0......absolute values cannot be negative.....  Did you enter your question correctly?

Jun 28, 2020