What are the solutions to the equation solvefor x,x^2-xy-1=0 ?

The solutions to the equation solvefor x,x^2-xy-1=0 are x=(y+sqrt(y^2+4))/2 ,x=(y-sqrt(y^2+4))/2

Find the zeros of solvefor x,x^2-xy-1=0

The zeros of solvefor x,x^2-xy-1=0 are x=(y+sqrt(y^2+4))/2 ,x=(y-sqrt(y^2+4))/2

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solve for

x, x^{2} - x y - 1 = 0

x = \frac{y + y ^{2} + 4}{2}, x = \frac{y - y ^{2} + 4}{2}

Solution steps

x^{2} - x y - 1 = 0

Write in the standard form

a x^{2} + b x + c = 0

x^{2} - y x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - y ) \pm ( - y ) ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

Simplify

(- y)^{2} - 4 \cdot 1 \cdot (- 1) : y^{2} + 4

x_{1, 2} = \frac{- ( - y ) \pm y ^{2} + 4}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - y ) + y ^{2} + 4}{2 \cdot 1}, x_{2} = \frac{- ( - y ) - y ^{2} + 4}{2 \cdot 1}

x = \frac{- ( - y ) + y ^{2} + 4}{2 \cdot 1} : \frac{y + y ^{2} + 4}{2}

x = \frac{- ( - y ) - y ^{2} + 4}{2 \cdot 1} : \frac{y - y ^{2} + 4}{2}

The solutions to the quadratic equation are:

x = \frac{y + y ^{2} + 4}{2}, x = \frac{y - y ^{2} + 4}{2}

5 - 8 c^{2} = - 75

x^{2} + 22 x + 101 = 0

- x^{2} + 3 x + 1 = - x + 1

16 + x^{2} = 49

2 x^{2} - 14 x = - 15 - x^{2}

What are the solutions to the equation solvefor x,x^2-xy-1=0 ?
The solutions to the equation solvefor x,x^2-xy-1=0 are x=(y+sqrt(y^2+4))/2 ,x=(y-sqrt(y^2+4))/2
Find the zeros of solvefor x,x^2-xy-1=0
The zeros of solvefor x,x^2-xy-1=0 are x=(y+sqrt(y^2+4))/2 ,x=(y-sqrt(y^2+4))/2