Solve w= (va)/(r^2+a^2) for a.
I applied the quadratic formula
a= (v/w +- sqrt (v^2/w^2 - 4*(-wr^2)) /2
and I have the answer as a= (v+/- sqrt[v^2+4w^2r^2]/w is this right?
No.
my quadratic was
wa^2 - va + wr^2 = 0
then a = (v ±√(v^2 - 4w^2r^2)/(2w)
You were close
Thanks
To solve the equation w = (va)/(r^2 + a^2) for a, you correctly applied the quadratic formula. However, there seems to be a small typo in your final answer.
Starting with the quadratic formula: a = (-b ± √(b^2 - 4ac)) / (2a), let's apply it to the equation w = (va)/(r^2 + a^2).
First, let's rewrite the equation in the standard quadratic form: a^2 + (vr^2/w)a - (vr^2/w) = 0.
Comparing this with the general quadratic equation form of ax^2 + bx + c = 0, we have:
a^2 + (vr^2/w)a - (vr^2/w) = 0.
Now, we can directly apply the quadratic formula:
a = (-b ± √(b^2 - 4ac)) / (2a).
Plugging in the values from our equation, we have:
a = (-(vr^2/w) ± √((vr^2/w)^2 - 4(1)(-vr^2/w))) / (2(1)).
Simplifying further:
a = (-vr^2 ± √((v^2r^4/w^2) + 4vr^2)) / (2).
Finally, you can simplify the square root term, leading to the correct final answer:
a = (-vr^2 ± √(v^2r^4/w^2 + 4vr^2)) / 2w.
So the correct answer to solve w = (va) / (r^2 + a^2) for a is: a = (-vr^2 ± √(v^2r^4/w^2 + 4vr^2)) / 2w.