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

The solutions to the equation 4+(x-2)^2=x(4-x) are x=2

Find the zeros of 4+(x-2)^2=x(4-x)

The zeros of 4+(x-2)^2=x(4-x) are x=2

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

x = 2

Solution steps

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

Expand

4 + (x - 2)^{2} : x^{2} - 4 x + 8

Expand

x (4 - x) : 4 x - x^{2}

x^{2} - 4 x + 8 = 4 x - x^{2}

Move

x^{2}

to the left side

2 x^{2} - 4 x + 8 = 4 x

Move

4 x

to the left side

2 x^{2} - 8 x + 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 2 \cdot 8}{2 \cdot 2}

(- 8)^{2} - 4 \cdot 2 \cdot 8 = 0

x_{1, 2} = \frac{- ( - 8 ) \pm 0}{2 \cdot 2}

x = \frac{- ( - 8 )}{2 \cdot 2}

\frac{- ( - 8 )}{2 \cdot 2} = 2

x = 2

The solution to the quadratic equation is:

x = 2

4.9 x^{2} - 30.88 x - 4 = 0

t^{2} - 6 t - 27 = 0

x^{2} + 14 x - 10 = 5

5 d (d + 1) + 6 = 0

x, x^{2} + 5 x - 6 = 0

What are the solutions to the equation 4+(x-2)^2=x(4-x) ?
The solutions to the equation 4+(x-2)^2=x(4-x) are x=2
Find the zeros of 4+(x-2)^2=x(4-x)
The zeros of 4+(x-2)^2=x(4-x) are x=2