What is the inverse of 8x+sqrt(9x^2+1)+1 ?

The inverse of 8x+sqrt(9x^2+1)+1 is (8x-8+sqrt(9x^2-18x+64))/(55),(8x-8-sqrt(9x^2-18x+64))/(55)

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inverse of 8x+sqrt(9x^2+1)+1

inverse

8 x + 9 x^{2} + 1 + 1

\frac{8 x - 8 + 9 x ^{2} - 18 x + 64}{55}, \frac{8 x - 8 - 9 x ^{2} - 18 x + 64}{55}

Solution steps

y = 8 x + 9 x^{2} + 1 + 1

Replace

x

with

y

x = 8 y + 9 y^{2} + 1 + 1

Solve for

y, x = 8 y + 9 y^{2} + 1 + 1

\frac{8 x - 8 + 9 x ^{2} - 18 x + 64}{55}, \frac{8 x - 8 - 9 x ^{2} - 18 x + 64}{55}

11 x^{3} - 10

\frac{1}{0.1 + j ^{0.4}}

\frac{6}{2 - 4 sin ( θ )}

\frac{x ^{2} - x - 12}{x ^{2} - 9}

What is the inverse of 8x+sqrt(9x^2+1)+1 ?
The inverse of 8x+sqrt(9x^2+1)+1 is (8x-8+sqrt(9x^2-18x+64))/(55),(8x-8-sqrt(9x^2-18x+64))/(55)