What are the solutions to the equation solvefor x,f(4+h)=x^2-2x ?

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

Find the zeros of solvefor x,f(4+h)=x^2-2x

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

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solve for

x, f (4 + h) = x^{2} - 2 x

x = 1 + f (h + 4) + 1, x = 1 - f (h + 4) + 1

Solution steps

f (4 + h) = x^{2} - 2 x

Expand

f (4 + h) : 4 f + f h

4 f + f h = x^{2} - 2 x

Switch sides

x^{2} - 2 x = 4 f + f h

Subtract

4 f + f h

from both sides

x^{2} - 2 x - (4 f + f h) = 4 f + f h - (4 f + f h)

Simplify

x^{2} - 2 x - 4 f - f h = 0

Solve with the quadratic formula

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

Simplify

(- 2)^{2} - 4 \cdot 1 \cdot (- 4 f - f h) : 2 1 + f (4 + h)

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

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 1 + f ( 4 + h )}{2 \cdot 1}, x_{2} = \frac{- ( - 2 ) - 2 1 + f ( 4 + h )}{2 \cdot 1}

x = \frac{- ( - 2 ) + 2 1 + f ( 4 + h )}{2 \cdot 1} : 1 + f (h + 4) + 1

x = \frac{- ( - 2 ) - 2 1 + f ( 4 + h )}{2 \cdot 1} : 1 - f (h + 4) + 1

The solutions to the quadratic equation are:

x = 1 + f (h + 4) + 1, x = 1 - f (h + 4) + 1

2 x^{2} + \frac{1}{2} = - 1

- y^{2} = 0

x^{2} + x - 0.1 = 0

x^{2} = 6 x

0.28 x^{2} - 8.8 x - 4.4 = 0

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