Question

The number of solutions (s) of the equation $\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2$ is

• A. one
• B. two
• C. four
• D. infinite

Answer 1

The correct option is A one

We can write,

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

$5x^2+10x+14=5(x^2+2x+1)+9=5(x+1{)}^2+9\ge 9$

L.H.S $\ge 2+3=5$

and R.H.S$=5-(x+1{)}^2\le 5$

Now, L.H.S$=$R.H.S needs $(x+1{)}^2=0$

so, $x=-1$ is only one solutions.