What are the solutions to the equation (3-x)^2+x^2=2(5-x) ?

The solutions to the equation (3-x)^2+x^2=2(5-x) are x=(2+sqrt(6))/2 ,x=(2-sqrt(6))/2

Find the zeros of (3-x)^2+x^2=2(5-x)

The zeros of (3-x)^2+x^2=2(5-x) are x=(2+sqrt(6))/2 ,x=(2-sqrt(6))/2

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(3 - x)^{2} + x^{2} = 2 (5 - x)

x = \frac{2 + 6}{2}, x = \frac{2 - 6}{2}

Decimal Notation

x = 2.22474 \dots, x = - 0.22474 \dots

Solution steps

(3 - x)^{2} + x^{2} = 2 (5 - x)

Expand

(3 - x)^{2} + x^{2} : 9 - 6 x + 2 x^{2}

Expand

2 (5 - x) : 10 - 2 x

9 - 6 x + 2 x^{2} = 10 - 2 x

Move

2 x

to the left side

2 x^{2} - 4 x + 9 = 10

Move

10

to the left side

2 x^{2} - 4 x - 1 = 0

Solve with the quadratic formula

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

(- 4)^{2} - 4 \cdot 2 (- 1) = 26

x_{1, 2} = \frac{- ( - 4 ) \pm 2 6}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 6}{2 \cdot 2}, x_{2} = \frac{- ( - 4 ) - 2 6}{2 \cdot 2}

x = \frac{- ( - 4 ) + 2 6}{2 \cdot 2} : \frac{2 + 6}{2}

x = \frac{- ( - 4 ) - 2 6}{2 \cdot 2} : \frac{2 - 6}{2}

The solutions to the quadratic equation are:

x = \frac{2 + 6}{2}, x = \frac{2 - 6}{2}

- 6 x^{2} - 8 (- 8 + 4) = 8

25 x^{2} + 18 x - 2 = 0

x^{2} + 15 x - 30 = 0

x^{2} - 3 (2 x - 5) = 10

x (2 - x) = - 2

What are the solutions to the equation (3-x)^2+x^2=2(5-x) ?
The solutions to the equation (3-x)^2+x^2=2(5-x) are x=(2+sqrt(6))/2 ,x=(2-sqrt(6))/2
Find the zeros of (3-x)^2+x^2=2(5-x)
The zeros of (3-x)^2+x^2=2(5-x) are x=(2+sqrt(6))/2 ,x=(2-sqrt(6))/2