What are the solutions to the equation (1+x)^2+x^2=11 ?

The solutions to the equation (1+x)^2+x^2=11 are x=(-1+sqrt(21))/2 ,x=-(1+sqrt(21))/2

Find the zeros of (1+x)^2+x^2=11

The zeros of (1+x)^2+x^2=11 are x=(-1+sqrt(21))/2 ,x=-(1+sqrt(21))/2

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(1 + x)^{2} + x^{2} = 11

x = \frac{- 1 + 21}{2}, x = - \frac{1 + 21}{2}

Decimal Notation

x = 1.79128 \dots, x = - 2.79128 \dots

Solution steps

(1 + x)^{2} + x^{2} = 11

Expand

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

1 + 2 x + 2 x^{2} = 11

Move

11

to the left side

2 x^{2} + 2 x - 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 2 ( - 10 )}{2 \cdot 2}

2^{2} - 4 \cdot 2 (- 10) = 221

x_{1, 2} = \frac{- 2 \pm 2 21}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 2 + 2 21}{2 \cdot 2}, x_{2} = \frac{- 2 - 2 21}{2 \cdot 2}

x = \frac{- 2 + 2 21}{2 \cdot 2} : \frac{- 1 + 21}{2}

x = \frac{- 2 - 2 21}{2 \cdot 2} : - \frac{1 + 21}{2}

The solutions to the quadratic equation are:

x = \frac{- 1 + 21}{2}, x = - \frac{1 + 21}{2}

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

5 t^{2} - 11 t = 12

x + 3 x + 1.5 x 44 x - 2 = 169

7 x^{2} + 54 x - 16 = 0

6 n^{2} + 40 = 70

What are the solutions to the equation (1+x)^2+x^2=11 ?
The solutions to the equation (1+x)^2+x^2=11 are x=(-1+sqrt(21))/2 ,x=-(1+sqrt(21))/2
Find the zeros of (1+x)^2+x^2=11
The zeros of (1+x)^2+x^2=11 are x=(-1+sqrt(21))/2 ,x=-(1+sqrt(21))/2