What is the inverse of sqrt(x^2-3x-10) ?

The inverse of sqrt(x^2-3x-10) is (3+sqrt(4x^2+49))/2 ,(3-sqrt(4x^2+49))/2

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inverse of sqrt(x^2-3x-10)

inverse

x^{2} - 3 x - 10

\frac{3 + 4 x ^{2} + 49}{2}, \frac{3 - 4 x ^{2} + 49}{2}

Solution steps

y = x^{2} - 3 x - 10

Replace

x

with

y

x = y^{2} - 3 y - 10

Solve for

y, x = y^{2} - 3 y - 10

\frac{3 + 4 x ^{2} + 49}{2}, \frac{3 - 4 x ^{2} + 49}{2}

f (x) = - (x - 1) - 1

- x^{2} - 2 x

y = - 2 \cdot (\frac{1}{3})^{x}

\frac{4 - x ^{2}}{x + 4}

f (x) = \frac{( 8 x - 2 )}{x + 5}

What is the inverse of sqrt(x^2-3x-10) ?
The inverse of sqrt(x^2-3x-10) is (3+sqrt(4x^2+49))/2 ,(3-sqrt(4x^2+49))/2