What is the answer to G(x)=-sqrt(25-x^2) ?

The answer to G(x)=-sqrt(25-x^2) is x= 5/(sqrt(G^2+1)) {G = 0}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

G(x)=-sqrt(25-x^2)

G (x) = - 25 - x^{2}

x = \frac{5}{G ^{2} + 1} {G \leq 0}, x = - \frac{5}{G ^{2} + 1} {G \geq 0}

Solution steps

G (x) = - 25 - x^{2}

Square both sides:

G^{2} x^{2} = 25 - x^{2}

G^{2} x^{2} = 25 - x^{2}

Solve

G^{2} x^{2} = 25 - x^{2} : x = \frac{5}{G ^{2} + 1}, x = - \frac{5}{G ^{2} + 1}

x = \frac{5}{G ^{2} + 1}, x = - \frac{5}{G ^{2} + 1}

Verify Solutions:

x = \frac{5}{G ^{2} + 1} {G \leq 0}, x = - \frac{5}{G ^{2} + 1} {G \geq 0}

The solutions are

x = \frac{5}{G ^{2} + 1} {G \leq 0}, x = - \frac{5}{G ^{2} + 1} {G \geq 0}

5 x + 3 + 5 x + 2 = 21

3 t + 4 = 3

y^{2} - 8 = y - 2

x^{2} + 44 = 2 4 x - 1

What is the answer to G(x)=-sqrt(25-x^2) ?
The answer to G(x)=-sqrt(25-x^2) is x= 5/(sqrt(G^2+1)) {G<= 0},x=-5/(sqrt(G^2+1)) {G>= 0}