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$$\begin{align} a^2 - b^2 = 3\\ a \cdot b = 2 \end{align}$$

In aforementioned equations, we can mentally find out the value of $a = 2, b = 1$. But what is the general way to solve this system algebraically?

I tried to use substitution but I got stuck. Rearranging two equations to $a^2=b^2+3$ and $a^2 \cdot b^2 = 2^2$ we will get:

$$ (b^2 + 3) \cdot b^2 = 4\\ \rightarrow b^4 + 3b^2 = 4\\ \rightarrow b^2(b^2 + 3) = 4\\ $$

Then what?

Alternatively can we solve by elimination?

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What you wrote is not a linear system. – heropup Mar 23 '14 at 4:35
@heropup you are right. I fixed it. – bman Mar 23 '14 at 4:36
Note that a=-2 and b=-1 is also a solution. – JB King Mar 23 '14 at 4:37
up vote 4 down vote accepted


$$\implies b^4+3b^2-4=0\iff (b^2+4)(b^2-1)=0$$

Then from the second equation, find corresponding value of $a$ for each value of $b$

Then, check whether each pair of $(a,b)$ satisfy the first

Alternatively, $$(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2$$

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Thank for thorough explanation. I just did not understand the alternative solution. Is it alternative way to solve the HINT section or it is totally another way for finding the solution? – bman Mar 23 '14 at 4:42
@Baqer, The second one is an independent method. – lab bhattacharjee Mar 23 '14 at 4:44
@Baqer, We have $$(a^2+b^2)^2=3^2+4^2=25\implies a^2+b^2=\pm5,$$ and we already have $$a^2-b^2$$ So, find $a^2,b^2$ and check for each pair of (a,b) if they satisfy the second equation – lab bhattacharjee Mar 23 '14 at 4:46

Your last equation, $b^2+3b^2-4=0$ is a quadratic in $b^2$. You can define $c=b^2$ and find $c^2+3c-4=0$. Maybe you can factor this, if not you can use the quadratic formula. Solve for $c$, then take the square root to get $b$. This works for quartics that only have the even power terms.

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