Question

if a+b+c=10,a-b-c=30 and abc=20, then what is the value of the triplet (a,b,c)?​

Answer 1

$\large{\text{\underline{To find:-}}$

The triplets $(a,b,c)$ that satisfy the system equation $\begin{cases} &a+b+c=10 \\ &a-b-c=30 \\ &abc=20 \end{cases}$.

$\large{\text{\underline{Solution:-}}$

We observe that the sum of the 1st and 2nd equation is $2a=40$. Hence, $a=20$.

Now the remaining system equation is,

$\hookrightarrow\begin{cases} &b+c=-10 \\ &bc=1 \end{cases}$

This equation can be solved by a quadratic equation. Consider a minimal polynomial that has, say $t=b,c$, as a solution.

$\hookrightarrow t^{2}+10t+1=0\implies\therefore t=-5\pm2\sqrt{6}$

Since $a.b$ are the solutions of the equation, we get,

$(a,b,c)=(20,-5+2\sqrt{6},-5-2\sqrt{6}), (20,-5-2\sqrt{6},-5+2\sqrt{6})$

$\large{\text{\underline{Conclusion:-}}$

There are two triplets of a, b, c, which are $\begin{cases} & a=20 \\ & b=-5\pm2\sqrt{6} \\ & c=-5\mp2\sqrt{6} \end{cases}$.