Question

which eq has no real solutions?
`2 = √(x + 3 + 5)`
`4 = √(x - 1 - 2)`
`1 = \sqrt[3]{x + 1 + 2}`
`6 = \sqrt[3]{x - 2 - 1}`

Answer 1

Let's solve each equation.

`2=\sqrt[]{x+3}+5`

First, we subtract 5 on each side.

`\begin{gathered} 2-5=\sqrt[]{x+3}+5-5 \\ -3=\sqrt[]{x+3} \end{gathered}`

You can observe that we've got x = 6 as a solution, however, this is not completely true because at the beginning we got a square root equal to a negative number and that doesn't have a solution in the real numbers. Square roots can't give a negative result, that's why.

The second equation is

`4=\sqrt[]{x-1}-2`

First, we add 2 on each side.

`\begin{gathered} 4+2=\sqrt[]{x-1}-2+2 \\ 6=\sqrt[]{x-1} \end{gathered}`

Then, we elevate the equation to the square power.

`\begin{gathered} 6^2=(\sqrt[]{x-1})^2 \\ 36=x-1 \\ x=36+1=37 \end{gathered}`

The second equation has a real solution.

The third equation is

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

The third equation has a real solution.

The fourth equation is

`\begin{gathered} 6=\sqrt[3]{x-2}-1 \\ 6+1=\sqrt[3]{x-2} \\ 7=\sqrt[3]{x-2} \\ 7^3=(\sqrt[3]{x-2})^3 \\ 343=x-2 \\ x=343+2=345 \end{gathered}`

The fourth equation has a real solution.