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$\sqrt{x}\sqrt{x}=4$ then $x=4$

But $\sqrt{x}\sqrt{x}=\sqrt{x^2}$

So, $\sqrt{x^2}=4$ which leads to $|x|=4$.

Why is this happening?

Michael
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user_
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  • I edited the question to put into latex mode (it makes the math symbols look nice). You can see how I did it by looking at the source. – Michael Aug 16 '19 at 18:15

2 Answers2

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The equality $\sqrt x\times \sqrt x=\sqrt {x^2}$ is valid only if $x\ge 0$; therefore$$\sqrt{x^2}=|x|=x=4$$

Mostafa Ayaz
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0

Before starting solving an equation with square roots, you have to put the C.E (condition of existance) of the radicants. In this case $x\geq0$, so the solution $x=-4$ isn't acceptable.

Matteo
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