What are the solutions to the equation (4-y)^2+y^2=10 ?

The solutions to the equation (4-y)^2+y^2=10 are y=3,y=1

Find the zeros of (4-y)^2+y^2=10

The zeros of (4-y)^2+y^2=10 are y=3,y=1

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(4 - y)^{2} + y^{2} = 10

y = 3, y = 1

Solution steps

(4 - y)^{2} + y^{2} = 10

Expand

(4 - y)^{2} + y^{2} : 16 - 8 y + 2 y^{2}

16 - 8 y + 2 y^{2} = 10

Move

10

to the left side

2 y^{2} - 8 y + 6 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 8 ) \pm ( - 8 ) ^{2} - 4 \cdot 2 \cdot 6}{2 \cdot 2}

(- 8)^{2} - 4 \cdot 2 \cdot 6 = 4

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

Separate the solutions

y_{1} = \frac{- ( - 8 ) + 4}{2 \cdot 2}, y_{2} = \frac{- ( - 8 ) - 4}{2 \cdot 2}

y = \frac{- ( - 8 ) + 4}{2 \cdot 2} : 3

y = \frac{- ( - 8 ) - 4}{2 \cdot 2} : 1

The solutions to the quadratic equation are:

y = 3, y = 1

(22 - x)^{2} + (11 - x)^{2} = x^{2}

5 y^{2} - 4 y - 2 = 0

x^{2} = - 5 x - 2

2 x^{2} + 2 x = - 3

32 = (4 x)^{2} - 4

What are the solutions to the equation (4-y)^2+y^2=10 ?
The solutions to the equation (4-y)^2+y^2=10 are y=3,y=1
Find the zeros of (4-y)^2+y^2=10
The zeros of (4-y)^2+y^2=10 are y=3,y=1