Solve e^3-1240 v^2

Question

Find the roots

v_{1} = - \frac{e 310 e}{620}, v_{2} = \frac{e 310 e}{620}

Alternative Form

v_{1} \approx - 0.127271, v_{2} \approx 0.127271

Evaluate

e^{3} - 1240 v^{2}

To find the roots of the expression,set the expression equal to 0

e^{3} - 1240 v^{2} = 0

Move the constant to the right-hand side and change its sign

- 1240 v^{2} = 0 - e^{3}

Removing 0 doesn't change the value,so remove it from the expression

- 1240 v^{2} = - e^{3}

Change the signs on both sides of the equation

1240 v^{2} = e^{3}

Divide both sides

\frac{1240 v ^{2}}{1240} = \frac{e ^{3}}{1240}

Divide the numbers

v^{2} = \frac{e ^{3}}{1240}

Take the root of both sides of the equation and remember to use both positive and negative roots

v = \pm \frac{e ^{3}}{1240}

Simplify the expression

More Steps

v = \pm \frac{e 310 e}{620}

Separate the equation into 2 possible cases

v = \frac{e 310 e}{620} v = - \frac{e 310 e}{620}

Solution

v_{1} = - \frac{e 310 e}{620}, v_{2} = \frac{e 310 e}{620}

Alternative Form

v_{1} \approx - 0.127271, v_{2} \approx 0.127271

Show Solution