Question

Please help me ASAP!!!

Answer 1

To solve the equation we first squared both sides of it, that is:

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

Now we can use the general formula for quadratic equations, then:

`\begin{gathered} x=\frac{-1\pm\sqrt[]{1^2-4(1)(-2)}}{2(1)} \\ =\frac{-1\pm\sqrt[]{1+8}}{2} \\ =\frac{-1\pm\sqrt[]{9}}{2} \\ =(-1\pm3)/(2) \\ \text{then} \\ x=(-1+3)/(2)=(2)/(2)=1 \\ or \\ x=(-1-3)/(2)=-(4)/(2)=-2 \end{gathered}`

Now that we found two option for x, we have to check which of them is really a solution for the original equation (we have to do this since we squared the original equation to get rid of the root), to do this we plug the values we found to see if the eqaution holds.

If x=1:

`\begin{gathered} \sqrt[]{1+3}=1+1 \\ \sqrt[]{4}=2 \\ 2=2 \end{gathered}`

hence x=1 is a solution for the original equation.

If x=-2:

`\begin{gathered} \sqrt[]{-2+3}=-2+1 \\ \sqrt[]{1}=-1 \\ 1=-1 \end{gathered}`

since this is not true, x=-2 is not a solution for the original equation.

Therefore, the solution for the equation is x=1.