Question

Solve the equation : +3=1+1

Answer 1

We have to solve the equation

`\sqrt[]{x+3}=x+1`

To do this we are going to take the second power in both sides of the equation. Then

`\begin{gathered} \sqrt[]{x+3}=x+1 \\ (\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 use the general formula for quadratic functions

`\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) \end{gathered}`

Then

`\begin{gathered} x=(-1+3)/(2)=(2)/(2)=1 \\ or \\ x=(-1-3)/(2)=-(4)/(2)=-2 \end{gathered}`

Now, in a rational equation we have to be careful; since we have to take the power to solve it, it appears that the equation has two solutions but, as we are about to see, this is not the case. What we have to do know is to plug the values we obtained in the original equation to see which one is in fact a solution.

Let's see if x=1 is a solution, pluggin the value in the equation we have

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

hence, x=1 is a solution of the equation.

Let's see if x=-2 is a solution:

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

but this is a contradiction, then x=-2 is not a solution.

Therefore the solution to the equation is x=1.