LorrieW658822 LorrieW658822
  • 03-11-2022
  • Mathematics
contestada

find x...........[tex]4 + \sqrt{(x + 1)} = 0[/tex]

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GustavO734893 GustavO734893
  • 03-11-2022

Due to the properties of the square root function,

[tex]\begin{gathered} \sqrt[]{(x+1)}\to\text{exists} \\ \Rightarrow x+1\ge0 \\ \Rightarrow x\ge-1 \\ \text{and} \\ \sqrt[]{y}\ge0 \end{gathered}[/tex]

Solving the equation,

[tex]\begin{gathered} 4+\sqrt[]{(x+1)}=0 \\ \Rightarrow\sqrt[]{(x+1)}=-4!!! \end{gathered}[/tex]

This is not possible since the square root is always greater or equal to zero; therefore, the equation has no solutions.

There is no real number such that multiplied by itself is equal to -4

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