How do you solve $sqrt(k+9)-sqrtk=sqrt3$ ?

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Answer 1 · 2017-05-14T10:53:55

$k = 3$

The feasibility conditions are

$k+9 ge 0$ and $k ge 0$ so $rArr k ge 0$

so by inspection, the solution is for $k=3$ because then

$sqrt(3+9)-sqrt3=sqrt3$ or

$sqrt(4 xx 3) = 2sqrt3=sqrt3+sqrt3$

How do you solve sqrt(k+9)-sqrtk=sqrt3sqrt(k+9)-sqrtk=sqrt3?