Find x such that $\sqrt {x-6} = x\sqrt{x-6}$

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Find all real values of $x$ such that

$$\sqrt {x-6} = x\sqrt{x-6}$$

I got $x = -6$ and $x = 1$, but how can that be because both of these values of $x$ give a negative under the radical.

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3 Answers

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Note that you can cancel the sqrt factor unless it is zero. Also $x\geqslant6$

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Your equation is of type $y=zy$, that is, multiplying by $z$ does not change $y$, then $y$ better be zero. In this case $y=\sqrt{x-6}$, so $x=6$.

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Working in the reals,

$x\sqrt{x-6} - \sqrt{x-6} = 0$

$(x-1)(\sqrt{x-6}) = 0$

By the Zero-product property

$x - 1 = 0, \sqrt{x-6} = 0$

$x = 1, x = 6$

But $x \geq 6$ since the square root cannot be negative, so $x = 6$ is the only solution.

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