What is the answer to solvefor y,sqrt(2y^2+1)+y-1=0 ?

The answer to solvefor y,sqrt(2y^2+1)+y-1=0 is y=-2,y=0

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solvefor y,sqrt(2y^2+1)+y-1=0

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

y = - 2, y = 0

Solution steps

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

Remove square roots

2 y^{2} + 1 = y^{2} - 2 y + 1

Solve

2 y^{2} + 1 = y^{2} - 2 y + 1 : y = - 2, y = 0

y = - 2, y = 0

Verify Solutions:

y = - 2

True

, y = 0

True

The solutions are

y = - 2, y = 0

3 (x + 1)^{\frac{2}{3}} = 12

x + 11 + 1 = x

x - 4 x + 4 = 3

3 x - 2 + 2 x + 2 + 1 = 0

lo g_{10} (x) + lo g_{10} (x - 1) = lo g_{10} (3 x + 12)

What is the answer to solvefor y,sqrt(2y^2+1)+y-1=0 ?
The answer to solvefor y,sqrt(2y^2+1)+y-1=0 is y=-2,y=0