How do you solve $x + 2 = \sqrt{7 x} + 2$ $x + 2 = sqrt(7 x) + 2$ ?

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Answer 1 · 2018-04-26T05:20:55

$x = 0$ $x=0$ and $x = 7$ $x=7$

$x + 2 = \sqrt{7 x} + 2$ $x+2=sqrt(7x)+2$

Subtract 2 on both sides

$\Rightarrow x = \sqrt{7 x}$ $=> x=sqrt(7x)$

Square Both Sides

$x^{2} = 7 x$ $x^2=7x$

divide by $x$ $x$ on both sides (assuming $x \neq 0$ $x != 0$ )

$\Rightarrow x = 7$ $=> x=7$

However, we must check if $x = 0$ $x=0$ is a solution

$0 + 2 = \sqrt{7 \cdot 0} + 2$ $0+2=sqrt(7*0)+2$

$\Rightarrow 2 = 0 + 2$ $=> 2=0+2$

This is true, so $x = 0$ $x=0$ is a solution!

Thus, $x = 0$ $x=0$ and $x = 7$ $x=7$

How do you solve x + 2 = sqrt(7 x) + 2x+2=√7x+2x + 2 = sqrt(7 x) + 2?