What is the answer to sqrt(x^2+b^2)+x(1-e)=x ?

The answer to sqrt(x^2+b^2)+x(1-e)=x is x= b/(sqrt(e^2-1)) {b>= 0},x=-b/(sqrt(e^2-1)) {b<= 0}

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sqrt(x^2+b^2)+x(1-e)=x

x^{2} + b^{2} + x (1 - e) = x

x = \frac{b}{e ^{2} - 1} {b \geq 0}, x = - \frac{b}{e ^{2} - 1} {b \leq 0}

Solution steps

x^{2} + b^{2} + x (1 - e) = x

Expand

x^{2} + b^{2} + x (1 - e) : x^{2} + b^{2} + x - e x

x^{2} + b^{2} + x - e x = x

Remove square roots

x^{2} + b^{2} = e^{2} x^{2}

Solve

x^{2} + b^{2} = e^{2} x^{2} : x = \frac{b}{e ^{2} - 1}, x = - \frac{b}{e ^{2} - 1}

x = \frac{b}{e ^{2} - 1}, x = - \frac{b}{e ^{2} - 1}

Verify Solutions:

x = \frac{b}{e ^{2} - 1} {b \geq 0}, x = - \frac{b}{e ^{2} - 1} {b \leq 0}

The solutions are

x = \frac{b}{e ^{2} - 1} {b \geq 0}, x = - \frac{b}{e ^{2} - 1} {b \leq 0}

x^{2} - 5 x - 2 = 3

22 - x = 2 + 3 x

x, x yz = 2

4 v - 31 + 6 = 5

A, 0.7 (A)^{0.3} = 5

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