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

Guest Aug 18, 2017
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#1
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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
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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