#1**+2 **

\((x-\sqrt{48})^2\leq\frac{49}{3}\)

I am sorry if I have interpreted your inequality incorrectly. I am fairly certain that the equation above is what you were attempting to convey.

\((x-\sqrt{48})^2\leq\frac{49}{3}\) | Undo the square by square rooting both sides of the inequality. |

\(x-\sqrt{48}\leq|\sqrt{\frac{49}{3}}|\) | Remember the square root rule that states that \(\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\). I like to think of it as "distributing" the square root into both the numerator and denominator. Also, the square root function always results in a negative and positive answer. That's why the absolute value sign is there. |

\(x-\sqrt{48}\leq|\frac{\sqrt{49}}{\sqrt{3}}|\) | Simplify the fraction on the right side of the equation. |

\(x-\sqrt{48}\leq|\frac{7}{\sqrt{3}}|\) | Let's rationalize the right hand side of the equation by multiplying the numerator and denominator by |

\(\frac{7}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}}=\frac{7\sqrt{3}}{3}\) | Reinsert this back into the inequality. |

\(x-\sqrt{48}\leq|\frac{7\sqrt{3}}{3}|\) | The absolute value results in 2 answers: the positive and negative. Let's split that up. |

\(x-\sqrt{48}\leq\pm\frac{7\sqrt{3}}{3}\) |

Now, we must evaluate each inequality separately. The one for the plus and the one for the minus. To make it easier, I'll solve both one at a time! I'll do the positive version first:

\(x-\sqrt{48}\leq\frac{7\sqrt{3}}{3}\) | Add \(\sqrt{48}\) to both sides. |

\(x\leq\frac{7\sqrt{3}}{3}+\sqrt{48}\) | First, I'll make\(\sqrt{48}\) into simplest radical form. |

\(\sqrt{48}=\sqrt{16*3}=\sqrt{16}\sqrt{3}=4\sqrt{3}\) | Ok, because this radical is in simplest radical form, reinsert it back into the equation. |

\(x\leq\frac{7\sqrt{3}}{3}+\frac{4\sqrt{3}}{1}\) | Put \(\frac{4\sqrt{3}}{1}\) into a common denominator so that it is possible to add them together. |

\(\frac{4\sqrt{3}}{1}*\frac{3}{3}=\frac{12\sqrt{3}}{3}\) | Now that this fraction has a common denominator, let's add them together. |

\(x\leq\frac{7\sqrt{3}}{3}+\frac{12\sqrt{3}}{3}\) | Add the fractions together. |

\(x\leq\frac{19\sqrt{3}}{3}\) | Time to solve the second part of the equation. |

Now, we have to solve for the other scenario, \(x-\sqrt{48}\geq-\frac{7\sqrt{3}}{3}\). Now, this is a tad difficult to explain, but the absolute value function does funny things to an inequality. The rule is that \(|f(x)|\leq a\Rightarrow f(x)\leq a\hspace{1mm}\text{and}\hspace{1mm}f(x)\geq -a\).

\(x-\sqrt{48}\geq-\frac{7\sqrt{3}}{3}\) | Solve and isolate x by adding \(\sqrt{48}\) to both sides. |

\(x\geq-\frac{7\sqrt{3}}{3}+\sqrt{48}\) | Put \(\sqrt{48}\) into simplest radical form and put in common denomator. Because I have already done this in the previous equation, I'll just skip. |

\(x\geq-\frac{7\sqrt{3}}{3}+\frac{12\sqrt{3}}{3}\) | Add the fractions together now. |

\(x\geq\frac{5\sqrt{3}}{3}\) | |

Now, put the solutions together.

\(\frac{5\sqrt{3}}{3}\leq x\leq \frac{19\sqrt{3}}{3}\)

TheXSquaredFactor
Jul 2, 2017

#1**+2 **

Best Answer

\((x-\sqrt{48})^2\leq\frac{49}{3}\)

I am sorry if I have interpreted your inequality incorrectly. I am fairly certain that the equation above is what you were attempting to convey.

\((x-\sqrt{48})^2\leq\frac{49}{3}\) | Undo the square by square rooting both sides of the inequality. |

\(x-\sqrt{48}\leq|\sqrt{\frac{49}{3}}|\) | Remember the square root rule that states that \(\sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}\). I like to think of it as "distributing" the square root into both the numerator and denominator. Also, the square root function always results in a negative and positive answer. That's why the absolute value sign is there. |

\(x-\sqrt{48}\leq|\frac{\sqrt{49}}{\sqrt{3}}|\) | Simplify the fraction on the right side of the equation. |

\(x-\sqrt{48}\leq|\frac{7}{\sqrt{3}}|\) | Let's rationalize the right hand side of the equation by multiplying the numerator and denominator by |

\(\frac{7}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}}=\frac{7\sqrt{3}}{3}\) | Reinsert this back into the inequality. |

\(x-\sqrt{48}\leq|\frac{7\sqrt{3}}{3}|\) | The absolute value results in 2 answers: the positive and negative. Let's split that up. |

\(x-\sqrt{48}\leq\pm\frac{7\sqrt{3}}{3}\) |

Now, we must evaluate each inequality separately. The one for the plus and the one for the minus. To make it easier, I'll solve both one at a time! I'll do the positive version first:

\(x-\sqrt{48}\leq\frac{7\sqrt{3}}{3}\) | Add \(\sqrt{48}\) to both sides. |

\(x\leq\frac{7\sqrt{3}}{3}+\sqrt{48}\) | First, I'll make\(\sqrt{48}\) into simplest radical form. |

\(\sqrt{48}=\sqrt{16*3}=\sqrt{16}\sqrt{3}=4\sqrt{3}\) | Ok, because this radical is in simplest radical form, reinsert it back into the equation. |

\(x\leq\frac{7\sqrt{3}}{3}+\frac{4\sqrt{3}}{1}\) | Put \(\frac{4\sqrt{3}}{1}\) into a common denominator so that it is possible to add them together. |

\(\frac{4\sqrt{3}}{1}*\frac{3}{3}=\frac{12\sqrt{3}}{3}\) | Now that this fraction has a common denominator, let's add them together. |

\(x\leq\frac{7\sqrt{3}}{3}+\frac{12\sqrt{3}}{3}\) | Add the fractions together. |

\(x\leq\frac{19\sqrt{3}}{3}\) | Time to solve the second part of the equation. |

Now, we have to solve for the other scenario, \(x-\sqrt{48}\geq-\frac{7\sqrt{3}}{3}\). Now, this is a tad difficult to explain, but the absolute value function does funny things to an inequality. The rule is that \(|f(x)|\leq a\Rightarrow f(x)\leq a\hspace{1mm}\text{and}\hspace{1mm}f(x)\geq -a\).

\(x-\sqrt{48}\geq-\frac{7\sqrt{3}}{3}\) | Solve and isolate x by adding \(\sqrt{48}\) to both sides. |

\(x\geq-\frac{7\sqrt{3}}{3}+\sqrt{48}\) | Put \(\sqrt{48}\) into simplest radical form and put in common denomator. Because I have already done this in the previous equation, I'll just skip. |

\(x\geq-\frac{7\sqrt{3}}{3}+\frac{12\sqrt{3}}{3}\) | Add the fractions together now. |

\(x\geq\frac{5\sqrt{3}}{3}\) | |

Now, put the solutions together.

\(\frac{5\sqrt{3}}{3}\leq x\leq \frac{19\sqrt{3}}{3}\)

TheXSquaredFactor
Jul 2, 2017