+2446 Well, one way to solve this problem is to find the value for a that satisfies the given equation: \(a^2+\frac{1}{a^2}=14\). Let's do that!
| \(a^2+\frac{1}{a^2}=14\) | Let's multiply both sides of the equation by a2 to get rid of the pesky fraction. |
| \(a^2*a^2+a^2*\frac{1}{a^2}=14*a^2\) | Simplify the left and right hand side of the equation. |
| \(a^4+1=14a^2\) | Subtract 14a^2 from both sides. |
| \(a^4-14a^2+1=0\) | Solve the 4th power equation using the substitution a^2=x. If a^2=x, then 14a^2=14x and a^4=x^2 |
| \(x^2-14x+1=0\) | We have converted this 4th power equation into just a quadratic. Now solve it by using any method. This isn't factorable, so I'll use completing the square. Subtract 1 on both sides. |
| \(x^2-14x=-1\) | The coefficient of the x^2 term is already 1, so we can go on to the next step of converting the left side to a perfect-square trinomial. |
| \(x^2-14x+(\frac{b}{2})^2=-1+(\frac{b}{2})^2\) | What does the b stand for? It stands for the coefficient of the linear term.=, -14. Since we are adding it one side, we must add it to the other. |
| \(x^2-14x+(\frac{-14}{2})^2=-1+(\frac{-14}{2})^2\) | Simplify both sides of the equation. |
| \(x^2-14x+49=48\) | The left hand side is now a perfect-square trinomial. Let's convert it now. |
| \((x+\frac{b}{2})^2=48\) | This is the form of what it will look like. Take the coefficient of the linear term, -14, and half it to get it into the desired form. |
| \((x-7)^2=48\) | Take the square root of both sides. |
| \(x-7=\pm\sqrt{48}\) | Remember that when you take the square root of something, it results in the positive and negative answer. Add 7 to both sides. |
| \(x=7\pm\sqrt{48}\) | Simplify the radical to its simplest terms. |
| \(\sqrt{48}=\sqrt{16*3}=\sqrt{16}\sqrt{3}=4\sqrt{3}\) | |
| \(x=7\pm4\sqrt{3}\) | I'm going to split these solutions into separate ones. |
Ok, now we must not forget that we were solving for a--not x. Let's substitute it in.
| \(x_1=7+4\sqrt{3}\) | \(x_2=7-4\sqrt{3}\) | Substitute a^2 back into the equation. |
| \(a^2=7+4\sqrt{3}\) | \(a^2=7-4\sqrt{3}\) | Take the square root of both sides. |
| \(a=\pm\sqrt{7+4\sqrt{3}}\) | \(a=\pm\sqrt{7-4\sqrt{3}}\) | Now, we have to simplify those square roots. It is possible. |
Now, I have a clever way of simplifying these radicals so that they are nicer. I'm going to do it to both individually.
| \(\sqrt{7+4\sqrt{3}}\) | To simplify this, I'm going to add another term. What you want to do is get another term containing the square root of 3. |
| \(\sqrt{7+4\sqrt{3}+\sqrt{3}^2-3}\) | You might notice something. I've added 2 more terms. Let's simplify inside the radical. Notice how I haven't actually changed the value of the radical. Rearrange the terms such that the terms are in degree. |
| \(\sqrt{\sqrt{3}^2+4\sqrt{3}+4}\) | This should look very similar to a quadratic equation except without a variable. This happens to be factorable, but we must break up the middle term. Let's replace √3 with b. |
| \(\sqrt{b^2+4b+4}\) |
How are we going to break up the middle term? We'll use the AC method.
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Now, in our case, b is the √3, so we'll have to replace that.
| \(\sqrt{b^2+2b+2b+4}\) | Remember that b was our temporary placeholder for √3, so let's replace it. Now, all we have done is split up the b-term such that we can start grouping. |
| \(\sqrt{\sqrt{3}^2+2\sqrt{3}+2\sqrt{3}+4}\) | Let's group these terms together. |
| \(\sqrt{(\sqrt{3}^2+2\sqrt{3})+(2\sqrt{3}+4)}\) | Find the GCF in each group and extract it. |
| \(\sqrt{\sqrt{3}(\sqrt{3}+2)+2(\sqrt{3}+2)}\) | Group together knowing the rule that \(a*c+b*c=c(a+b)\). |
| \(\sqrt{(\sqrt{3}+2)(\sqrt{3}+2)}\) | Look what we have done! We have manipulated the answer into a perfect square. |
| \(\sqrt{(\sqrt{3}+2)(\sqrt{3}+2)}=\sqrt{(\sqrt{3}+2)^2}=2+\sqrt{3}\) | |
Do the same process with the other answer, \(\sqrt{7-4\sqrt{3}}\). Now, you could go through that cumbersome process again, but if you realize that it is the same thing with just a subtraction sign, it should be clear that the only change should be the negative sign when you solve it. And indeed, this is true. \(\sqrt{7-4\sqrt{3}}=2-\sqrt{3}\)
Let's separate the final values for a:
| \(a_1=+(2+\sqrt{3})\) | \(a_2=-(2+\sqrt{3})\) | \(a_3=+(2-\sqrt{3})\) | \(a_4=-(2-\sqrt{3})\) | Eval- uate all. |
| \(a_1=2+\sqrt{3}\) | \(a_2=-2-\sqrt{3}\) | \(a_3=+2-\sqrt{3}\) | \(a_4=-2+\sqrt{3}\) | |
Now, evaluate each and see which is largest.
Attempt 1:
| \(a^3+\frac{1}{a^3}\) | Plug in the first value for a. |
| \((2+\sqrt{3})^3+\frac{1}{(2+\sqrt{3})^3}\) | Apply distribution rule that states that \((a+b)^3=a^3+3a^2b+3ab^2+b^3\) |
| \((2+\sqrt{3})^3=2^3+3*2^2*\sqrt{3}+3*2*\sqrt{3}^2+\sqrt{3}^3\) | Simplify this. |
| \(8+12\sqrt{3}+18+3\sqrt{3}\) | Combine like terms. |
| \(26+15\sqrt{3}\) | Reinsert this into the original equation. |
| \(26+15\sqrt{3}+\frac{1}{26+15\sqrt{3}}*\frac{26-15\sqrt{3}}{26-15\sqrt{3}}\) | Rationalize the denominator by multiplying the denominator by (26-15√3). Simplify the numerator and denominator. |
| \((26+15\sqrt{3})*(26-15\sqrt{3})\) | Utilize the rule that \((a+b)(a-b)=a^2-b^2\) |
| \(26^2-(15\sqrt{3})^2\) | |
| \(576-(15\sqrt{3})^2\) | "Distribute" the exponent to every term in the multiplication because \((ab)^2=a^2b^2\) |
| \(15^2*\sqrt{3}^2=225*3=575\) | Reinsert. |
| \(576-575=1\) | Wow, the denominator is 1. That's useful. Let's get back to the original expression. |
| \(26+15\sqrt{3}+26-15\sqrt{3}\) | Combine like terms. |
| \(52\) | |
Do this same process for the other values for a.
a1=52
a2=-52
a3=52
a4=-52
Therefore, the largest possible value for a when plugged in is 52.
