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

The solutions to the equation solvefor x,f(x+1)=2x^2 are x=(f+sqrt(f^2+8f))/4 ,x=(f-sqrt(f^2+8f))/4

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

The zeros of solvefor x,f(x+1)=2x^2 are x=(f+sqrt(f^2+8f))/4 ,x=(f-sqrt(f^2+8f))/4

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solve for

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

x = \frac{f + f ^{2} + 8 f}{4}, x = \frac{f - f ^{2} + 8 f}{4}

Solution steps

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

Expand

f (x + 1) : x f + f

x f + f = 2 x^{2}

Switch sides

2 x^{2} = x f + f

Move

f

to the left side

2 x^{2} - f = x f

Move

x f

to the left side

2 x^{2} - f - x f = 0

Write in the standard form

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

2 x^{2} - f x - f = 0

Solve with the quadratic formula

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

Simplify

(- f)^{2} - 4 \cdot 2 (- f) : f^{2} + 8 f

x_{1, 2} = \frac{- ( - f ) \pm f ^{2} + 8 f}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - f ) + f ^{2} + 8 f}{2 \cdot 2}, x_{2} = \frac{- ( - f ) - f ^{2} + 8 f}{2 \cdot 2}

x = \frac{- ( - f ) + f ^{2} + 8 f}{2 \cdot 2} : \frac{f + f ^{2} + 8 f}{4}

x = \frac{- ( - f ) - f ^{2} + 8 f}{2 \cdot 2} : \frac{f - f ^{2} + 8 f}{4}

The solutions to the quadratic equation are:

x = \frac{f + f ^{2} + 8 f}{4}, x = \frac{f - f ^{2} + 8 f}{4}

r^{2} - 10 r = - 17

- 9 x^{2} + 1 = 0

x^{2} = 1225

x, P (x - Q)^{2} + R = 0

7 x^{2} - 14 x - 56 = 0

What are the solutions to the equation solvefor x,f(x+1)=2x^2 ?
The solutions to the equation solvefor x,f(x+1)=2x^2 are x=(f+sqrt(f^2+8f))/4 ,x=(f-sqrt(f^2+8f))/4
Find the zeros of solvefor x,f(x+1)=2x^2
The zeros of solvefor x,f(x+1)=2x^2 are x=(f+sqrt(f^2+8f))/4 ,x=(f-sqrt(f^2+8f))/4