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

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

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G(x)=sqrt(x+1)

G (x) = x + 1

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

Solution steps

G (x) = x + 1

Square both sides:

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

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

Solve

G^{2} x^{2} = x + 1 : 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}}

True

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

The solutions are

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

x - 2 = - x + 2

x + 3 x + 10 = 6

1.19 = \frac{d}{2}

10 + 24 k = 13 1 + k^{2}

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