What are the solutions to the equation x^2+(x+1)^2=41 ?

The solutions to the equation x^2+(x+1)^2=41 are x=4,x=-5

Find the zeros of x^2+(x+1)^2=41

The zeros of x^2+(x+1)^2=41 are x=4,x=-5

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x^{2} + (x + 1)^{2} = 41

x = 4, x = - 5

Solution steps

x^{2} + (x + 1)^{2} = 41

Expand

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

2 x^{2} + 2 x + 1 = 41

Move

41

to the left side

2 x^{2} + 2 x - 40 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 2 ( - 40 )}{2 \cdot 2}

2^{2} - 4 \cdot 2 (- 40) = 18

x_{1, 2} = \frac{- 2 \pm 18}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 2 + 18}{2 \cdot 2}, x_{2} = \frac{- 2 - 18}{2 \cdot 2}

x = \frac{- 2 + 18}{2 \cdot 2} : 4

x = \frac{- 2 - 18}{2 \cdot 2} : - 5

The solutions to the quadratic equation are:

x = 4, x = - 5

2 x + 5 + 1 > 4

x^{2} + 4 x = 3

a^{3} - 3 + 3 a^{2} - a

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

13 x^{2} - 38 x - 3

What are the solutions to the equation x^2+(x+1)^2=41 ?
The solutions to the equation x^2+(x+1)^2=41 are x=4,x=-5
Find the zeros of x^2+(x+1)^2=41
The zeros of x^2+(x+1)^2=41 are x=4,x=-5