Question

Sum of values of $x$, satisfying the equation $\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2,$ is

• A. $-3$
• B. $-1$
• C. $0$
• D. $3$

Answer 1

The correct option is B $-1$

We have,
$\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2$
$\Rightarrow \sqrt{3(x+1{)}^2+4}+\sqrt{5(x+1{)}^2+9}=5-(x+1{)}^2$

The minimum value of $\sqrt{3(x+1{)}^2+4}$ is 2 at $x=-1$ and the minimum value of $\sqrt{5(x+1{)}^2+9}$ is 3 at $x=-1$.
So, the minimum value of LHS is 5 at $x=-1$.
RHS has maximum value 5 at $x=-1$.

So the given equation is satisfied only at $x=-1$