+0

0
156
2

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

Guest Aug 18, 2017
Sort:

#1
+18827
+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
+1602
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

### 18 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details