+0  
 
0
197
6
avatar+56 

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
Sort: 

6+0 Answers

 #1
avatar
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
avatar+1337 
0

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

TheXSquaredFactor  Jul 9, 2017
 #2
avatar+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 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 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
avatar+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
avatar
0

Very elegant, Alan! Thank you.

Guest Jul 9, 2017
 #6
avatar+56 
0

Thank you guys so much!!!!

Trinityvamp286  Jul 9, 2017

11 Online Users

avatar
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