Given that $ab + c^2 = 18$ and $a^2 + b^2 = 12$, Find $abc$

Question

I have this question in my test.

Known that: $$ab + c^2 = 18$$ $$a^2 + b^2 = 12$$ Find the value of $a$$b$$c$

Can anyone give me a hint or a guide on what should I do? I don't need a straight answer. Thanks!

Answer 1

The answer is not unique. Take $$a = 0, b = \sqrt{12}, c = \sqrt{18}$$ and $$a = b = \sqrt{6}, c = \sqrt{12}.$$

Both answers satisfy the equations, but for the first solution we have $abc=0$, while in the second solution, $abc = 12\sqrt{3}$.

Answer 2

Let $b=ka$. Then

$$ka^2+c^2=18\tag1$$ $$(1+k^2)a^2=12\iff a^2=\frac{12}{1+k^2}\tag2$$

From $(1),(2)$ $$\frac{12k}{1+k^2}+c^2=18\implies c^2=18-\frac{12k}{1+k^2}\tag3$$

Thus

$$abc=ka^2c=\pm k\left(\frac{12}{1+k^2}\right)\sqrt{18-\frac{12k}{1+k^2}}\tag4$$

for any constant $k$.

Answer 3

From $a^2+b^2=12$ we take $\sin\theta=\dfrac{a}{\sqrt{12}}$ and $\cos\theta=\dfrac{b}{\sqrt{12}}$ so $$c^2=18-ab=18-12\sin\theta\cos\theta=6(3-\sin2\theta)$$ the value of $abc$ is $$f(\theta)=12\sin\theta\cos\theta\sqrt{6(3-\sin2\theta)}$$

Answer 4

Let $a=k$, where $-2\sqrt{3}\leq k\leq 2\sqrt{3}$. Then $b^2=12-k^2$ and hence $b=\pm\sqrt{12-k^2}$.

It follows that $c^2=18\mp k\sqrt{12-k^2}$ or $c=\pm\sqrt{18\mp k\sqrt{12-k^2}}$.

Since the question did not state any condition of $a, b, c$, there will be infinitely many solutions for the product $abc$