What is the answer to solvefor p,s(p)=sqrt(p)-p ?

The answer to solvefor p,s(p)=sqrt(p)-p is p= 1/(s^2+2s+1) {s -1},p=0

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solvefor p,s(p)=sqrt(p)-p

solve for

p, s (p) = p - p

p = \frac{1}{s ^{2} + 2 s + 1} {s < - 1 or s > - 1}, p = 0

Solution steps

s (p) = p - p

Remove square roots

p^{2} s^{2} + 2 p^{2} s + p^{2} = p

Solve

p^{2} s^{2} + 2 p^{2} s + p^{2} = p : p = \frac{1}{s ^{2} + 2 s + 1}, p = 0

p = \frac{1}{s ^{2} + 2 s + 1}, p = 0

Verify Solutions:

p = \frac{1}{s ^{2} + 2 s + 1} {s < - 1 or s > - 1}, p = 0

True

The solutions are

p = \frac{1}{s ^{2} + 2 s + 1} {s < - 1 or s > - 1}, p = 0

x - 4 = - 2

4 x + 8 + 13 = 19

b = 1.69 s + 4.45 - 3.49

What is the answer to solvefor p,s(p)=sqrt(p)-p ?
The answer to solvefor p,s(p)=sqrt(p)-p is p= 1/(s^2+2s+1) {s<-1\lor s>-1},p=0