How do you solve $sqrt(x+9)=sqrt3+sqrtx$ and check your solution?

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Answer 1 · 2016-10-15T14:45:34

$x= 3$

square both sides:

$(sqrt(x + 9))^2 = (sqrt(3) + sqrt(x))^2$

$x + 9 = 3 + x + 2sqrt(3x)$

$6 = 2sqrt(3x)$

Square again to get rid of the square root on the right.

$6^2 = (2sqrt(3x))^2$

$36 = 4(3x)$

$36 = 12x$

$x =3$

Finally, don't forget to check your solution in the original equation.

$sqrt(9 + 3) =^? sqrt(3) + sqrt(3)$

$sqrt(12) =^? 2sqrt(3)$

$sqrt(4 xx 3) =^? 2sqrt(3)$

$2sqrt(3) = 2sqrt(3)$

Hopefully this helps!

How do you solve sqrt(x+9)=sqrt3+sqrtxsqrt(x+9)=sqrt3+sqrtx and check your solution?