What are the solutions to the equation (1+y)^2+y^2=25 ?

The solutions to the equation (1+y)^2+y^2=25 are y=3,y=-4

Find the zeros of (1+y)^2+y^2=25

The zeros of (1+y)^2+y^2=25 are y=3,y=-4

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(1 + y)^{2} + y^{2} = 25

y = 3, y = - 4

Solution steps

(1 + y)^{2} + y^{2} = 25

Expand

(1 + y)^{2} + y^{2} : 1 + 2 y + 2 y^{2}

1 + 2 y + 2 y^{2} = 25

Move

25

to the left side

2 y^{2} + 2 y - 24 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 2 ( - 24 )}{2 \cdot 2}

2^{2} - 4 \cdot 2 (- 24) = 14

y_{1, 2} = \frac{- 2 \pm 14}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- 2 + 14}{2 \cdot 2}, y_{2} = \frac{- 2 - 14}{2 \cdot 2}

y = \frac{- 2 + 14}{2 \cdot 2} : 3

y = \frac{- 2 - 14}{2 \cdot 2} : - 4

The solutions to the quadratic equation are:

y = 3, y = - 4

x^{2} - 50 x + 225 = 0

(2 x + 3)^{2} - 14 x - 21 = - 6

x, y^{4} = 4 x^{2} + 6 x y

x, x^{2} + k x + 4 = 0

x^{2} + 8 x + 14 = 23

What are the solutions to the equation (1+y)^2+y^2=25 ?
The solutions to the equation (1+y)^2+y^2=25 are y=3,y=-4
Find the zeros of (1+y)^2+y^2=25
The zeros of (1+y)^2+y^2=25 are y=3,y=-4