Finding the values of $a$ and $b$

Question

We are given two equations:

$$ \left\{ \begin{array}{c} \sqrt{a}+b=11 \\ a+\sqrt{b}=7 \\ \end{array} \right. $$

How to find the value of $a$ and $b$ by solving the equation? All I could do was use the hit and trial method. Help me with this!

Answer 1

Perhaps this is what you already did, but I think it's easy enough to be a good method.

There aren't very many possibilities to try. a & b must be non-negative because 7 and 11 are both real. Also a & b must be perfect squares. So from the 1st equation b<=9. From the second a<=4. So a could be 0, 1, 4. a=0 and a=1 don't work by just plugging them in to the equations. So a should be 4. Which you can try and see that it works when b=9.

Answer 2

It's usually better to work with squares than square roots, since we wind up with polynomial equations to solve. Assuming $a,b \ge 0$ so square roots are real:

$$ a = (11 - b)^2 $$

$$ b = (7 - a)^2 $$

Substituting for $b$ in first equation from the second:

$$ a = (11 - (7-a)^2)^2 $$

gives a quartic equation to solve for (nonnegative) roots $a \in [0,7]$.

The polynomial equation turns out to be:

$$ a^4 - 28a^3 + 282a^2 - 1065a + 1444 = 0 $$

By Rational Roots Thm. any rational root would be an integer, and indeed there is an integer root in the desired range $[0,7]$. Since such $a$ will divide $1444$, there aren't too many possibilities to check.

After you factor that root out, we're left with an irreducible cubic polynomial, and by Descartes Rule of Signs it has at least one positive real root:

$$ a^3 - 24a^2 + 176a - 361 = 0 $$

one of which turns out to also be in the desired range, $a \approx 3.4156$. However this gives a value for $b = (7-a)^2$ which is greater than $11$, $(7-a)^2 \approx 12.8479$. In other words this is an "artifact" root created by the process of squaring both variables.

The only real solution of the original system of equations is $a=4,b=9$.