Find the number of real solution(s)

[anemone]

How many real solution(s) does the following system have?

$a+b=2$

$ab-c^2=1$

 

[kaliprasad]

How many real solution(s) does the following system have?

$a+b=2$

$ab-c^2=1$

Spoiler

$a+b=2$

$ab-c^2=1$

let $a = 1 -x$ and $b = 1 + x$
so $ab-c^2 = 1- x^2 - c^2 = 1$
so $x^2 + c^2 =0$
so $x = 0, c = 0$

so solution (a,b,c) = (1,1,0) or one solution

 

[juantheron]

How many real solution(s) does the following system have?

$a+b=2$

$ab-c^2=1$

Spoiler

Solution:: Given $ab=1+c^2\geq 0$. So $ab\geq 0$

So either $a,b\geq 0$ or $a,b\leq 0,$ but given $a+b=2$ .So $a,b\geq 0$

Now using $\bf{A.M\geq G.M},$ we get $\displaystyle \frac{a+b}{2}\geq \sqrt{ab}$

So $\displaystyle 1\geq \sqrt{ab}\Rightarrow ab\leq 1\Rightarrow 1+c^2\leq 1\Rightarrow c^2\leq 0$

So $c^2\leq 0\Rightarrow c^2 = 0\Rightarrow c = 0$(because Square Quantity $\geq 0$)

and equality hold when $a = b$. So Using $a+b=1$, we get $a = b = 1.$

So $(a,b,c) = (1,1,0)$