Solve for the positive value of x such that $\sqrt[3]{x^2 - 4x + 4} = 16$.

Guest Aug 18, 2017

#1**+2 **

Solve for the positive value of x such that \sqrt[3]{x^2 - 4x + 4} = 16

(\(\sqrt[3]{x^2 - 4x + 4}\) ) .

Formula:

\(\begin{array}{|rcll|} \hline && \sqrt[n]{a^m} \\ &=& (a^m)^\frac{1}{n} \\ &=& a^{m\cdot \frac{1}{n}} \\ &=& a^{\frac{m}{n}} \\ &=& a^{\frac{1}{n}\cdot m } \\ &=& (a^{\frac{1}{n}})^m \\ &=& (\sqrt[n]{a})^m \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \sqrt[3]{x^2 - 4x + 4} &=& 16 \quad & | \quad x^2 - 4x + 4 = (x-2)^2 \\ \sqrt[3]{(x-2)^2} &=& 16 \quad & | \quad \sqrt[n]{a^m} = (\sqrt[n]{a})^m \\ \left(~ \sqrt[3]{x-2} ~\right)^2 &=& 16 \quad & | \quad \sqrt{} \\ \sqrt[3]{x-2} &=& 4 \quad & | \quad cube \\ x-2 &=& 4^3 \\ x-2 &=& 64 \\ x &=& 64+2 \\ \mathbf{x} & \mathbf{=} & \mathbf{66} \\ \hline \end{array}\)

heureka
Aug 18, 2017

#2**0 **

Wait! There is another solution, as well! Heureka was so close to the second solution, too. I think the user forgot about taking the square root always results in a positive and negative answer. Anyway, I like Heureka's method, but here is my method of solving.

The original equation is \(\sqrt[3]{x^2-4x+4}=16\). Now let's solve for *x*:

\(\sqrt[3]{x^2-4x+4}=16\) | Raise both sides of the equation to third power to eliminate the cube root. |

\(x^2-2x+4=16^3\) | Do 16^3. luckily for us, we need not know the exact value of it yet. We can manipulate it in a base of 2 so we do not need to do a difficult calculation. |

\(x^2-2x+4=16^3\) | Usually, you would subtract 16^3 on both sides , but \(x^2-2x+4\) happens to be a perfect-square trinomial, and transforming it into one is much easier computationally. |

\((x-2)^2=16^3\) | Take the square root of both sides. Of course, when you take the square root of both sides, it results in a positive and a negative answer. |

\(|x-2|=\sqrt{16^3}\) | Now, I will use power rules creatively to simplify 16^3. |

\(|x-2|=\sqrt{(2^4)^3}\) | 2^4 is equal to 16, so I have not changed the right-hand side of the equation at all. Now, I will utilize another power rule that states that \((a^b)^c=a^{b*c}\) |

\(|x-2|=\sqrt{2^{4*3}}=\sqrt{2^{12}}\) | When you are taking the square root of an even power, just divide the power by 2. Let me show you why. |

\(\sqrt{a^{2k}}=\sqrt{(a^k)^2}=a^k\) | What I have shown here is that any number raised to the power of a number that is a multiple of 2 is simply halved when the square root is taken to it. |

\(\sqrt{2^{12}}=\sqrt{(2^6)^2}=2^6\) | I have done the exact same process as above, just with the numbers that are given. |

\(|x-2|=2^6\) | With absolute value expressions, the answer is divided into the positive and negative answer. Before we do that, however, we must evaluate 2^6. You may have memorized it, but I have a trick if you haven't. |

\(2^6=(2^3)^2=8^2=64\) | Therefore, 2^6=64. |

\(|x-2|=64\) | Now, let's solve for each equation separately. |

\(x-2=64\) | \(-(x-2)=64\) | |

\(x=66\) | \(x-2=-64\) | |

\(x=-62\) | ||

Normally, with equations inside radicals, you would have to check both solutions, but here it isn't necessary. This is because after cubing both sides, you end up with a quadratic, and an answer from a quadratic is always right--assuming no arithmetic error was made when solving. Therefore, your solution set is:

\(x_1=-62\)

\(x_2=66\)

TheXSquaredFactor
Aug 18, 2017