What is the inverse of f(x)=x^2+bx+1 ?

The inverse of f(x)=x^2+bx+1 is (-b+sqrt(b^2-4(1-x)))/2 ,(-b-sqrt(b^2-4(1-x)))/2

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inverse of f(x)=x^2+bx+1

inverse

\frac{- b + b ^{2} - 4 ( 1 - x )}{2}, \frac{- b - b ^{2} - 4 ( 1 - x )}{2}

Solution steps

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

Replace

x

with

y

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

Solve for

y, x = y^{2} + b y + 1

\frac{- b + b ^{2} - 4 ( 1 - x )}{2}, \frac{- b - b ^{2} - 4 ( 1 - x )}{2}

What is the inverse of f(x)=x^2+bx+1 ?
The inverse of f(x)=x^2+bx+1 is (-b+sqrt(b^2-4(1-x)))/2 ,(-b-sqrt(b^2-4(1-x)))/2