Relative error and roots of significant digits

Question

a.) The maximal relative error of the volume of a ball is allowed within $1\%$. What is the maximal relative error measuring the radius of the ball?

b.) Given the equation $x^2-40x+1=0$, find its roots to five significant digits using $\sqrt{399} = 19.975$, correctly rounded to five digits.

What I did and know for a is that the volume of a sphere is $V = 4/3\pi * R^3$ and the only term that can have a relative error is the R term. So, I let $dV$ be the relative error of the volume and $dR$ the relative error of the radius then

$dV = dR + dR + dR = 3dR$

$dV/3 = dR$

$1\%/3 = dR$

and I am completely lost in b.

Answer 1

Solve for $x$ over the real numbers:
$x^2-40 x+1 = 0$
Solve the quadratic equation by completing the square.
Subtract $1$ from both sides:
$x^2-40 x = -1$
Take one half of the coefficient of $x$ and square it, then add it to both sides.
Add $400$ to both sides:
$x^2-40 x+400 = 399$
Factor the left hand side.
Write the left hand side as a square:
$(x-20)^2 = 399$
Eliminate the exponent on the left hand side.
Take the square root of both sides:
$x-20 = \sqrt{(399)} \text{ or } x-20 = -\sqrt{(399)}$
Look at the first equation: Solve for $x$.
Add $20$ to both sides:
$x = 20+\sqrt{(399)} \text{ or } x-20 = -\sqrt{(399)}$
Look at the second equation: Solve for x.
Add $20$ to both sides:
Answer:
$$ x = 20+\sqrt{(399)} \text{ or } x = 20-\sqrt{(399)} $$

For question a): $$ V = \frac{4}{3} \pi R^3 \Rightarrow dV = 4\pi R^2 dR \Rightarrow \frac{dV}{V} = \frac{4\pi R^2dR}{\frac{4}{3} \pi R^3} \le .01 \Rightarrow \frac{3dR}{R} \le .01 \Rightarrow \frac{dR}{R} \le \frac{1}{300} \Rightarrow \max{\left(\frac{dR}{R}\right)} = \frac{1}{300} $$