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

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!

## closed as off-topic by Dietrich Burde, Claude Leibovici, user91500, John B, Behrouz MalekiFeb 3 '17 at 13:03

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Dietrich Burde, Claude Leibovici, user91500, John B, Behrouz Maleki
If this question can be reworded to fit the rules in the help center, please edit the question.

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, $abc = 12\sqrt{3}$.

• Why the solution isn't unique.? – Nosrati Jan 27 '17 at 13:08
• @MyGlasses because there are 3 variables and two equations. – Turambar Jan 27 '17 at 13:22

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 $k$.

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 $$f(\theta)=12\sin\theta\cos\theta\sqrt{6(3-\sin2\theta)}$$

• A shorter form would be $$f(\theta)=6 \sqrt{6} \sin 2\theta \sqrt{3-\sin2\theta}$$ – Yuriy S Jan 27 '17 at 13:36
• @YuriyS The minimum of $abc$ is $0$ and the maximum.? – Nosrati Jan 27 '17 at 13:39
• MyGlasses, maximum is $12 \sqrt{3}$ – Yuriy S Jan 27 '17 at 13:44

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 $abc$