What is the answer to ((-2x^2+5x-1))/(2x-1)=0 ?

The answer to ((-2x^2+5x-1))/(2x-1)=0 is x=(5-sqrt(17))/4 ,x=(5+sqrt(17))/4

Solve for x: ((-2x^2+5x-1))/(2x-1)=0

The solution is x=(5-sqrt(17))/4 ,x=(5+sqrt(17))/4

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\frac{( - 2 x ^{2} + 5 x - 1 )}{2 x - 1} = 0

x = \frac{5 - 17}{4}, x = \frac{5 + 17}{4}

Solution steps

\frac{( - 2 x ^{2} + 5 x - 1 )}{2 x - 1} = 0

Simplify

\frac{( - 2 x ^{2} + 5 x - 1 )}{2 x - 1} : \frac{- 2 x ^{2} + 5 x - 1}{2 x - 1}

\frac{- 2 x ^{2} + 5 x - 1}{2 x - 1} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

- 2 x^{2} + 5 x - 1 = 0

Solve

- 2 x^{2} + 5 x - 1 = 0 : x = \frac{5 - 17}{4}, x = \frac{5 + 17}{4}

x = \frac{5 - 17}{4}, x = \frac{5 + 17}{4}

Verify Solutions

Find undefined (singularity) points:

x = \frac{1}{2}

Combine undefined points with solutions:

x = \frac{5 - 17}{4}, x = \frac{5 + 17}{4}

16 x^{- 4} - 17 x^{- 2} + 1 = 0

\frac{7}{x} = \frac{20}{5}

\frac{x}{x + 4} + 1 = \frac{16}{x + 4}

\frac{2}{x} = (1.2)

\frac{1}{x + 1} = 2

What is the answer to ((-2x^2+5x-1))/(2x-1)=0 ?
The answer to ((-2x^2+5x-1))/(2x-1)=0 is x=(5-sqrt(17))/4 ,x=(5+sqrt(17))/4
Solve for x: ((-2x^2+5x-1))/(2x-1)=0
The solution is x=(5-sqrt(17))/4 ,x=(5+sqrt(17))/4