What is the answer to G(x)=sqrt(1-x) ?

The answer to G(x)=sqrt(1-x) is x=(-1+sqrt(1+4G^2))/(2G^2) {G 0},x=(-1-sqrt(1+4G^2))/(2G^2)

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G(x)=sqrt(1-x)

G (x) = 1 - x

x = \frac{- 1 + 1 + 4 G ^{2}}{2 G ^{2}} {G < 0 or G > 0}, x = \frac{- 1 - 1 + 4 G ^{2}}{2 G ^{2}}

Solution steps

G (x) = 1 - x

Square both sides:

G^{2} x^{2} = 1 - x

G^{2} x^{2} = 1 - x

Solve

G^{2} x^{2} = 1 - x : x = \frac{- 1 + 1 + 4 G ^{2}}{2 G ^{2}}, x = \frac{- 1 - 1 + 4 G ^{2}}{2 G ^{2}}

x = \frac{- 1 + 1 + 4 G ^{2}}{2 G ^{2}}, x = \frac{- 1 - 1 + 4 G ^{2}}{2 G ^{2}}

Verify Solutions:

x = \frac{- 1 + 1 + 4 G ^{2}}{2 G ^{2}} {G < 0 or G > 0}, x = \frac{- 1 - 1 + 4 G ^{2}}{2 G ^{2}}

True

The solutions are

x = \frac{- 1 + 1 + 4 G ^{2}}{2 G ^{2}} {G < 0 or G > 0}, x = \frac{- 1 - 1 + 4 G ^{2}}{2 G ^{2}}

3 x - 5 - x + 7 = 1

2 = x

x + x^{2} = 0

x, 2 x - 5 y = - 1

3 x + 1 = 0

What is the answer to G(x)=sqrt(1-x) ?
The answer to G(x)=sqrt(1-x) is x=(-1+sqrt(1+4G^2))/(2G^2) {G<0\lor G>0},x=(-1-sqrt(1+4G^2))/(2G^2)