Question

The number of solutions to the equation $\sqrt{6-4x-x^2}=x+4$ is:

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

The correct option is C

$2$

Step 1: Given data.

It is given that the equation is $\sqrt{6-4x-x^2}=x+4$.

Now, squaring on both sides, we get:

$\begin{array}{rcl}{\left(\sqrt{6-4x-x^2}\right)}^2& =& {\left(x+4\right)}^2\\ 6-4x-x^2& =& x^2+8x+16\\ 2x^2+12x+10& =& 0\end{array}$

Step 2: Factorise the equation $2x^2+12x+10=0$.

The equation $2x^2+12x+10=0$ is in the form of a quadratic equation. So factorize the equation as:

$\begin{array}{rcl}2x^2+12x+10& =& 0\\ x^2+6x+5& =& 0\\ x^2+x+5x+5& =& 0\\ x\left(x+1\right)+5\left(x+1\right)& =& 0\\ \left(x+1\right)\left(x+5\right)& =& 0\\ x& =& -1,x=-5\end{array}$

Hence, the solution of the given equation is $x=-5,-1$ . Therefore, the number of solutions for the given equation is $2$. Thus, option C is correct.