Simultaneous Equations $a^2 - b^2 = -16$ and $2ab = 30$

Question

I need help solving these simultaneous equations:

$$a^2 - b^2 = -16$$ $$2ab = 30$$

Answer 1

After dividing by $2$ we get $$b=\frac{15}{a}$$ plugging this in the first equation we get $$a^2-\left(\frac{15}{a}\right)^2=-16$$ Can you proceed? After multiplying by $$a^2$$ we obtain $$a^4+16a^2-225=0$$ now substitute $$a^2=t$$

Answer 2

Hint:

You may use the polarisation identity: $$xy=\tfrac14\bigl( (x+y)^2-(x-y)^2\bigr)$$ to deduce $a^2+b^2$. This will ultimately produce a linear systemin $a^2$ and $b^2$.

Answer 3

Oftentimes this particular type of simultaneous equations (which often arise when finding square roots of complex numbers for example) have "nice" whole number solutions. It's a good idea to look for these whole number solutions first by inspection before doing any long calculations.

In this case, we know from the second equation that $ab = 15$. So if we want whole number solutions, we should try with $a,b$ being factors of 15 (so just $\{1,15\}$ or $\{ 3,5\}$ here (or negatives of these)). By looking at the first equation, we see by inspection that $a=3,b=5$ will work (since $3\times 5=15$ and $3^2-5^2=9-25=-16$). Hence we obtain the solution $a=3,b=5$. Also $a=-3,b=-5$ will be the other solution.

(If this question was asking for finding the square root of a complex number, once you found one solution, say $a=3,b=5$, you could immediately write down that the square roots are $\pm(3+5i)$. If the question just came on it's own, you would probably need to provide explanation of why these are all the solutions.)

Answer 4

I looked for an integer solution:

$$2ab = 30 \implies ab = 15$$ $$a\ and\ b \in \{\pm1,\pm3,\pm5,\pm15\}$$ Trial and error yields $$\{a,b\}\ = \{-3,-5\}\ or\ \{3,5\}$$