Question

How many real solutions does the equation V2 = I + 1 have?OA. 0OB.1OC.2OD.cannot be determined from the graph

Answer 1

Try to find x from the equality

$x+1=\sqrt[]{x-2}$$\begin{gathered} (x+1)^2=x-2 \\ x^2+2x+1=x-2 \\ x^2+2x-x+1+2=0 \\ x^2+x+3=0 \end{gathered}$

we can use this equation to solve x

$x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}$

were a =1, b=1 and c=3

so replacing

$\begin{gathered} x=\frac{-1\pm\sqrt[]{1^2-4(1)(3)}}{2(1)} \\ \\ x=\frac{-1\pm\sqrt[]{1-12}}{2} \\ \\ x=\frac{-1\pm\sqrt[]{-11}}{2} \end{gathered}$

the negative number within the root indicates that the solution is imaginary

so the solution can not be determined from the graph