What is the answer to G(x)=sqrt(x^2-9) ?

The answer to G(x)=sqrt(x^2-9) is x=sqrt(-9/(G^2-1)) {0<= G<1},x=-sqrt(-9/(G^2-1)) {-1<G<= 0}

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G(x)=sqrt(x^2-9)

G (x) = x^{2} - 9

x = - \frac{9}{G ^{2} - 1} {0 \leq G < 1}, x = - - \frac{9}{G ^{2} - 1} {- 1 < G \leq 0}

Solution steps

G (x) = x^{2} - 9

Square both sides:

G^{2} x^{2} = x^{2} - 9

G^{2} x^{2} = x^{2} - 9

Solve

G^{2} x^{2} = x^{2} - 9 : x = - \frac{9}{G ^{2} - 1}, x = - - \frac{9}{G ^{2} - 1}

x = - \frac{9}{G ^{2} - 1}, x = - - \frac{9}{G ^{2} - 1}

Verify Solutions:

x = - \frac{9}{G ^{2} - 1} {0 \leq G < 1}, x = - - \frac{9}{G ^{2} - 1} {- 1 < G \leq 0}

The solutions are

x = - \frac{9}{G ^{2} - 1} {0 \leq G < 1}, x = - - \frac{9}{G ^{2} - 1} {- 1 < G \leq 0}

x^{2} + 16 = 30

x + 1 = 10

x - 7 + 1 = x

3 = 5 - x

What is the answer to G(x)=sqrt(x^2-9) ?
The answer to G(x)=sqrt(x^2-9) is x=sqrt(-9/(G^2-1)) {0<= G<1},x=-sqrt(-9/(G^2-1)) {-1<G<= 0}