What are the solutions to the equation 4x^2=(x+1)^2+b ?

The solutions to the equation 4x^2=(x+1)^2+b are x=-(-1+sqrt(3b+4))/3 ,x=(sqrt(3b+4)+1)/3

Find the zeros of 4x^2=(x+1)^2+b

The zeros of 4x^2=(x+1)^2+b are x=-(-1+sqrt(3b+4))/3 ,x=(sqrt(3b+4)+1)/3

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4 x^{2} = (x + 1)^{2} + b

x = - \frac{- 1 + 3 b + 4}{3}, x = \frac{3 b + 4 + 1}{3}

Solution steps

4 x^{2} = (x + 1)^{2} + b

Expand

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

4 x^{2} = x^{2} + 2 x + 1 + b

Switch sides

x^{2} + 2 x + 1 + b = 4 x^{2}

Move

4 x^{2}

to the left side

- 3 x^{2} + 2 x + b + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 3 ) ( b + 1 )}{2 ( - 3 )}

Simplify

2^{2} - 4 (- 3) (b + 1) : 2 3 b + 4

x_{1, 2} = \frac{- 2 \pm 2 3 b + 4}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- 2 + 2 3 b + 4}{2 ( - 3 )}, x_{2} = \frac{- 2 - 2 3 b + 4}{2 ( - 3 )}

x = \frac{- 2 + 2 3 b + 4}{2 ( - 3 )} : - \frac{- 1 + 3 b + 4}{3}

x = \frac{- 2 - 2 3 b + 4}{2 ( - 3 )} : \frac{3 b + 4 + 1}{3}

The solutions to the quadratic equation are:

x = - \frac{- 1 + 3 b + 4}{3}, x = \frac{3 b + 4 + 1}{3}

y, (x - 1)^{2} + y^{2} = 0

x^{2} + 16 x + 71

x^{2} - 109 = - 8 x

x^{2} = 1 2^{2} + 6^{2}

- 16 x^{2} + 73 x + 6 = 0

What are the solutions to the equation 4x^2=(x+1)^2+b ?
The solutions to the equation 4x^2=(x+1)^2+b are x=-(-1+sqrt(3b+4))/3 ,x=(sqrt(3b+4)+1)/3
Find the zeros of 4x^2=(x+1)^2+b
The zeros of 4x^2=(x+1)^2+b are x=-(-1+sqrt(3b+4))/3 ,x=(sqrt(3b+4)+1)/3