What are the solutions to the equation solvefor x,p(x+1)=x^2+p ?

The solutions to the equation solvefor x,p(x+1)=x^2+p are x=p,x=0

Find the zeros of solvefor x,p(x+1)=x^2+p

The zeros of solvefor x,p(x+1)=x^2+p are x=p,x=0

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solve for

x, p (x + 1) = x^{2} + p

x = p, x = 0

Solution steps

p (x + 1) = x^{2} + p

Expand

p (x + 1) : x p + p

x p + p = x^{2} + p

Switch sides

x^{2} + p = x p + p

Move

p

to the left side

x^{2} = x p

Move

x p

to the left side

x^{2} - x p = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} - p x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - p ) \pm ( - p ) ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

Simplify

(- p)^{2} - 4 \cdot 1 \cdot 0 : p

x_{1, 2} = \frac{- ( - p ) \pm p}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - p ) + p}{2 \cdot 1}, x_{2} = \frac{- ( - p ) - p}{2 \cdot 1}

x = \frac{- ( - p ) + p}{2 \cdot 1} : p

x = \frac{- ( - p ) - p}{2 \cdot 1} : 0

The solutions to the quadratic equation are:

x = p, x = 0

a, u = 50 a - 0.5 a^{2} + 60 m - 0.25 m^{2}

(a - b) \cdot x^{2} + (b + c - a) x - c = 0

(\frac{3}{5} x - \frac{2}{3})^{2} = \frac{5}{6}

u, 1 + d^{2} u^{2} = (1 + \frac{d ^{2}}{2})^{2}

4 m^{2} + 9 m + 2 = 0

What are the solutions to the equation solvefor x,p(x+1)=x^2+p ?
The solutions to the equation solvefor x,p(x+1)=x^2+p are x=p,x=0
Find the zeros of solvefor x,p(x+1)=x^2+p
The zeros of solvefor x,p(x+1)=x^2+p are x=p,x=0