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

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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 ?

Trinityvamp286  Jul 9, 2017
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#1
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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

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)

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)

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
+1337
0

Well, Im glad after all that work that we both get the same answers!

TheXSquaredFactor  Jul 9, 2017
#2
+1337
0

Well, one way to solve this problem is to find the value for 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.

\                /

\              /

\     4   /

\         /

\       /

\    /

2     \/    2

/\

/  \

/     \

/        \

/     4     \

/              \

Now, in our case, 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.

TheXSquaredFactor  Jul 9, 2017
#4
+26322
+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
#5
0

Very elegant, Alan! Thank you.

Guest Jul 9, 2017
#6
+56
0

Thank you guys so much!!!!

Trinityvamp286  Jul 9, 2017

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