What is the inverse of y=sqrt(2x^2-5x) ?

The inverse of y=sqrt(2x^2-5x) is (5+sqrt(25+8x^2))/4 ,(5-sqrt(25+8x^2))/4

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inverse of y=sqrt(2x^2-5x)

inverse

y = 2 x^{2} - 5 x

\frac{5 + 25 + 8 x ^{2}}{4}, \frac{5 - 25 + 8 x ^{2}}{4}

Solution steps

y = 2 x^{2} - 5 x

Replace

x

with

y

x = 2 y^{2} - 5 y

Solve for

y, x = 2 y^{2} - 5 y

\frac{5 + 25 + 8 x ^{2}}{4}, \frac{5 - 25 + 8 x ^{2}}{4}

5 x^{2} + 5 x

f (x) = 3 x^{2} - 4.03

\frac{x - 3}{x ^{2} - 3 x - 18}

\frac{6 + 2 x}{6 - 2 x}

y = 4 x - 3

What is the inverse of y=sqrt(2x^2-5x) ?
The inverse of y=sqrt(2x^2-5x) is (5+sqrt(25+8x^2))/4 ,(5-sqrt(25+8x^2))/4