suppose a is a real number for which a^2+1/a^2=14 ...

what is the largest possible value of a^3+1/a^3 ?

Answer quickly please? Explanation as well.. Thankyou

Trinityvamp286 Jul 9, 2017

#1**0 **

Solve for a:

a^2 + 1/a^2 = 14

Bring a^2 + 1/a^2 together using the common denominator a^2:

(a^4 + 1)/a^2 = 14

Multiply both sides by a^2:

a^4 + 1 = 14 a^2

Subtract 14 a^2 from both sides:

a^4 - 14 a^2 + 1 = 0

Substitute x = a^2:

x^2 - 14 x + 1 = 0

Subtract 1 from both sides:

x^2 - 14 x = -1

Add 49 to both sides:

x^2 - 14 x + 49 = 48

Write the left hand side as a square:

(x - 7)^2 = 48

Take the square root of both sides:

x - 7 = 4 sqrt(3) or x - 7 = -4 sqrt(3)

Add 7 to both sides:

x = 7 + 4 sqrt(3) or x - 7 = -4 sqrt(3)

Substitute back for x = a^2:

a^2 = 7 + 4 sqrt(3) or x - 7 = -4 sqrt(3)

Take the square root of both sides:

a = sqrt(7 + 4 sqrt(3)) or a = -sqrt(7 + 4 sqrt(3)) or x - 7 = -4 sqrt(3)

7 + 4 sqrt(3) = 4 + 4 sqrt(3) + 3 = 4 + 4 sqrt(3) + (sqrt(3))^2 = (sqrt(3) + 2)^2:

a = sqrt(3) + 2 or a = -sqrt(7 + 4 sqrt(3)) or x - 7 = -4 sqrt(3)

7 + 4 sqrt(3) = 4 + 4 sqrt(3) + 3 = 4 + 4 sqrt(3) + (sqrt(3))^2 = (sqrt(3) + 2)^2:

a = 2 + sqrt(3) or a = -sqrt(3) + 2 or x - 7 = -4 sqrt(3)

Add 7 to both sides:

a = 2 + sqrt(3) or a = -2 - sqrt(3) or x = 7 - 4 sqrt(3)

Substitute back for x = a^2:

a = 2 + sqrt(3) or a = -2 - sqrt(3) or a^2 = 7 - 4 sqrt(3)

Take the square root of both sides:

a = 2 + sqrt(3) or a = -2 - sqrt(3) or a = sqrt(7 - 4 sqrt(3)) or a = -sqrt(7 - 4 sqrt(3))

7 - 4 sqrt(3) = 4 - 4 sqrt(3) + 3 = 4 - 4 sqrt(3) + (sqrt(3))^2 = (2 - sqrt(3))^2:

a = 2 + sqrt(3) or a = -2 - sqrt(3) or a = 2 - sqrt(3) or a = -sqrt(7 - 4 sqrt(3))

7 - 4 sqrt(3) = 4 - 4 sqrt(3) + 3 = 4 - 4 sqrt(3) + (sqrt(3))^2 = (2 - sqrt(3))^2:

**Answer: | a = 2 + sqrt(3) or a = -2 - sqrt(3) or a = 2 - sqrt(3) or a = -2 - sqrt(3)**

**If you sub: (2+ sqrt(3))^3 + 1/(2+ sqrt(3))^3 =52**

Guest Jul 9, 2017

#3

#2**0 **

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 ato get rid of the pesky fraction. ^{2 } |

\(a^2*a^2+a^2*\frac{1}{a^2}=14*a^2\) | 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.

\ /

\ /

\ **4** /

\ /

\ /

\ /

**2** \/ ** 2**

/\

/ \

/ \

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/ **4** \

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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*.

*a*_{1}=52

a_{2}=-52

a_{3}=52

a_{4}=-52

Therefore, the largest possible value for a when plugged in is 52.

TheXSquaredFactor Jul 9, 2017

#4**+1 **

Alternative approach:

a^2 + 1/a^2 = 14. (1)

From (1): a^2 = 14 - 1/a^2. Multiply through by a to get: a^3 = 14a - 1/a. (2)

From (1): 1/a^2 = 14 - a^2. Divide through by a to get: 1/a^3 = 14/a - a. (3)

Add (2) and (3): a^3 + 1/a^3 = 13(a + 1/a). (4)

Now (a + 1/a)^2 = a^2 + 1/a^2 + 2 → 14 + 2 → 16 using (1)

Hence a + 1/a = 4. (5)

Put (5) into (4): a^3 + 1/a^3 = 13*4 → 52

.

Alan Jul 9, 2017