Question

$I:$ $\sqrt{2x+7}+\sqrt{x+4}=0$ has real roots.
$II:$ $\sqrt{6-4x-x^2}=x+4$ has $-1$ only as the root. Which of the above statement(s) is/are true ?

• A. only $I$
• B. only $II$
• C. both $I$ and $II$
• D. neither $I$ nor $II$

Answer 1

The correct option is B only $II$

Equation 1: $\sqrt{2x+7}+\sqrt{x+4}=0$

Sum of two positive real numbers can not be zero.Therefore statement 1 is not true.

Equation 2 can be written as $6-4x-x^2=(x+4{)}^2$

$\Rightarrow x^2+6x+5=0$

$\Rightarrow (x+1)(x+5)=0$

$\Rightarrow x=(-1,-5)$

Putting $x=-1$, the LHS becomes $\sqrt{6+4-1}=+3$ or $-3$

RHS: $-1+4=+3$

$\therefore$ The statement is true.