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How do you solve $2p = sqrtp+5$ ?

1 Answer

Konstantinos Michailidis

Oct 6, 2015

Set $p=t^2>=0$ hence we have that

$2t^2=sqrt(t^2)+5=>2t^2=t+5=>2t^2-t-5=0=> t=1/4(1+sqrt41)$

hence $p=t^2=1/8(21+sqrt41)$

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