Question

Help solve the question please.

Answer 1

Answer:

$v = \sqrt{\dfrac{a -cb}{c\sqrt{2}} } = \dfrac{\sqrt{ a -cb}}{\sqrt{c}\sqrt[4]{2} }$

Step-by-step explanation:

$c = \dfrac{a}{b+\sqrt{2v} }$

The equation for $v$ will be

$c = \dfrac{a}{b+\sqrt{2v} } \implies c(b+\sqrt{2v}) = a \implies cb + c\sqrt{2v} = a$

Once $\sqrt{2v} = \sqrt{2} \sqrt{v}$

$cb + c\sqrt{2}\sqrt{v} = a \implies c\sqrt{2}\sqrt{v} = a -cb \implies \sqrt{v} =\dfrac{a -cb}{c\sqrt{2}}$

Square both sides

$\sqrt{v} =\dfrac{a -cb}{c\sqrt{2}} \implies v = \sqrt{\dfrac{a -cb}{c\sqrt{2}} } = \dfrac{\sqrt{ a -cb}}{\sqrt{c}\sqrt[4]{2} }$

Answer 2

Step-by-step explanation:

Multiply both sides of the equation by $b + \sqrt{2\nu}$ to get

$cb + c\sqrt{2\nu} = a$

Put the cb term to the right hand side and then divide by c to get

$\sqrt{2\nu} = \dfrac{a - cb}{c}$

Taking the square of the equation, we get

$2\nu = \left(\dfrac{a - cb}{c}\right)^2$

Finally, dividing the equation by 2 and we get an equation for $\nu.$

$\nu = \dfrac{1}{2}\left(\dfrac{a - cb}{c}\right)^2$