What are the solutions to the equation (b+1)^2+b^2=(b+9)^2 ?

The solutions to the equation (b+1)^2+b^2=(b+9)^2 are b=20,b=-4

Find the zeros of (b+1)^2+b^2=(b+9)^2

The zeros of (b+1)^2+b^2=(b+9)^2 are b=20,b=-4

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(b + 1)^{2} + b^{2} = (b + 9)^{2}

b = 20, b = - 4

Solution steps

(b + 1)^{2} + b^{2} = (b + 9)^{2}

Expand

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

Expand

(b + 9)^{2} : b^{2} + 18 b + 81

2 b^{2} + 2 b + 1 = b^{2} + 18 b + 81

Move

81

to the left side

2 b^{2} + 2 b - 80 = b^{2} + 18 b

Move

18 b

to the left side

2 b^{2} - 16 b - 80 = b^{2}

Move

b^{2}

to the left side

b^{2} - 16 b - 80 = 0

Solve with the quadratic formula

b_{1, 2} = \frac{- ( - 16 ) \pm ( - 16 ) ^{2} - 4 \cdot 1 \cdot ( - 80 )}{2 \cdot 1}

(- 16)^{2} - 4 \cdot 1 \cdot (- 80) = 24

b_{1, 2} = \frac{- ( - 16 ) \pm 24}{2 \cdot 1}

Separate the solutions

b_{1} = \frac{- ( - 16 ) + 24}{2 \cdot 1}, b_{2} = \frac{- ( - 16 ) - 24}{2 \cdot 1}

b = \frac{- ( - 16 ) + 24}{2 \cdot 1} : 20

b = \frac{- ( - 16 ) - 24}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

b = 20, b = - 4

4 x^{2} + 2 x = 6

5 x - x^{2} = 6

F, g b + N b = 2 F^{2}

c^{2} + 2 c = 48

6 r + r^{2} = 27

What are the solutions to the equation (b+1)^2+b^2=(b+9)^2 ?
The solutions to the equation (b+1)^2+b^2=(b+9)^2 are b=20,b=-4
Find the zeros of (b+1)^2+b^2=(b+9)^2
The zeros of (b+1)^2+b^2=(b+9)^2 are b=20,b=-4