Question

The number of solution of $\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

As $\sqrt{3x^2+6x+7}=\sqrt{3{(x+1)}^2+4}\ge 2,$

and $\sqrt{5x^2+10x+14}=\sqrt{5{(x+1)}^2+9}\ge 3$
$\therefore L.H.S.\ge 5$ and always positive

Now

$R.H.S.=4-2x-x^2=5-{(x+1)}^2\le 5$
$\therefore$ The equation holds only when, $L.H.S.=R.H.S.=5$
$i.e.4-2x-x^2=5$

$\Rightarrow x^2+2x+1=0$

$\Rightarrow x=-1$

Hence, only one solution to the above equation.