$3-sqrt(3n²+n-3)=n+1$ how do you solve this equation ?

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Answer 1 · 2017-05-14T10:37:21

$n = (-7/2,1)$

$3-sqrt(3n²+n-3)=n+1rArr sqrt(3n²+n-3)=2-n$

now squaring both sides

$3n²+n-3=4-4n+n^2$

simplifying

$2 n^2 + 5 n - 7=0$

and solving for $n$

$n = (-7/2,1)$ both are feasible solutions because for them $3n²+n-3 > 0$

3-sqrt(3n²+n-3)=n+13-sqrt(3n²+n-3)=n+1 how do you solve this equation ?