Equation with two unknowns

Question

I have this equation to solve and I solved it but don't know if the result is correct.

$\begin{cases}2\pi r_1+2\pi r_2=24 & (1) \\ \pi r_1^2+\pi r_2^2=20 & (2)\end{cases}$

Equation $(1)$ gives $r_1=\frac{12-\pi r_2}{\pi}$

Plug that into $(2)$, we have $\pi\big(\frac{12-\pi r_2}{\pi}\big)^2+\pi r_2^2=20 \Rightarrow 2\pi r_2^2-24r_2+\frac{144}{\pi}-20=0$.

$\frac {2\pi r_2^2-24r_2+\frac{144}{\pi}-20}{2\pi}=0$

So to solve the unknowns I use the reduced quadratic equation

$r_2^2 - 3.819r_2 + 4.11 = 0$

$r = - \frac {- 3.819}{2} {+ \choose -} \sqrt{\frac {- 3.819^2}{2^2}-4.11}$

$r = 1.9 {+ \choose -} \sqrt{-0.463}$

So because I don't have complex numbers in my course I think that this equation does not have an solution.

I'm I correct in my assumption or is my assumption wrong?

Thanks!!

Answer 1

Let's consider the system \begin{cases} x+y=a\\ x^2+y^2=b \end{cases} which is the same as yours with $a=12/\pi$ and $b=20/\pi$. We can rewrite the second equation as $$ (x+y)^2-2xy=b $$ so, taking into account the first equation, it becomes $$ 2xy=a^2-b $$ Now the problem is reduced to finding two numbers of which we know the sum and the product: they are the roots of $$ z^2-az+\frac{a^2-b}{2}=0 $$ The discriminant is $$ a^2-4\frac{a^2-b}{2}=a^2-2a^2+2b=2b-a^2 $$ so we have real solutions if and only if $$ 2b-a^2\ge0 $$ In your case $$ 2\frac{20}{\pi}-\frac{144}{\pi^2}=\frac{40\pi-144}{\pi^2} $$ and $144/40=3.6>\pi$, so the discriminant is negative.

Answer 2

935485
$\begin{vmatrix} 2\pi&2\pi r_1-24&0\\ 0&2\pi&2\pi r_1-24\\ \pi&0&\pi r^2_1-20\\ \end{vmatrix}=0$ is a quadratic equation derived from your two equations. When solved, it yields the two values for $r_1$. Substitute these two values into your equation (1) and solve to get the corresponding values for $r_2$.