If x^4 + x^(-4) = 7, what is the value of x^8 + x^(-8) =?

Question

A. 14
B. 21
C. 47
D. 49
E. 51

Bunuel · Answer

Square  x^4 + x^{(-4)} = 7 :

x^8 + 2*x^4*x^{(-4)} + x^{(-8)} = 49 .

Notice that  x^4*x^{(-4)} = x^4*\frac{1}{x^4} = 1 .

So we have  x^8 + 2*1 + x^{(-8)} = 49 .

Thus  x^8 + x^{(-8)} = 47 .

Answer: C.

LaveenaPanchal · Answer

x^4 + x ^-4 = 7
x^4 + 1/x^4 = 7

Squaring on both sides we get, x^8 + 2(x^8)(1/x^8)+1/x^8 = 49
x^8+x^-8 = 49-2 = 47

elevateV · Answer

Given,
x^4 + x ^-4 = 7
(consider a = x^4 and b= x ^-4
if we square both the sides, (a+b)^2 = a*a + b*b + 2*a*b
here, 2*a*b will cancel out x term )
Squaring both the sides
x^8 + 2(x^8)(1/x^8)+1/x^8 = 49
x^8 + 1/x^8 = 47

Answer (C)

CrackverbalGMAT · Answer

This is a simple question where you have to apply a basic algebraic identity to obtain the value of the given expression. 
 (a+b)^2  =  a^2 + b^2 + 2ab . 
Therefore,  a^2 + b^2 = (a+b)^2 – 2ab

x^4 + x^{(-4)}  = 7, can be rewritten as  x^4 + \frac{1}{x^4}  = 7.

x^8 + x^{(-8)}  =  x^8 + \frac{1}{x^8} . Observe that  x^8 = (x^4)^2 ; similarly,  \frac{1}{x^8} = (\frac{1}{x^4})^2 .

So, we are essentially trying to find the value of an expression in the form of  a^2 + b^2 .

Therefore,  x^8 + \frac{1}{x^8}  =   ^2 – 2*x^4*\frac{1}{x^4}  
 =   ^2  – 2 {since the  x^4  cancels out}
 = 49 – 2 = 47.

The correct answer option is C.

Hope that helps!
Aravind B T

v12345 · Answer

x^4 + \frac{1}{x^4} = 7

Squaring,
 x^8 + \frac{1}{x^8} + 2 = 49

x^8 + \frac{1}{x^8} = 47

Hence, OA is C.

MathRevolution · Answer

=> Let  x^4 = A  , then  x^{-4} = \frac{1}{x^4} = \frac{1}{A}

=>  x^4 + x^{-4} = 7

=>  A + \frac{1}{A} = 7

=>  A^2 + \frac{1}{A^2} + 2 = 49  (On squaring both the sides)

=>  A^2 + \frac{1}{A^2} = 49 - 2 = 47

=>  x^8 = (x^4)^2 = A^2

=>  x^8 + \frac{1}{(x^8)}= A^2 + \frac{1}{A^2} = 47

Answer C

bumpbot · Answer

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If x^4 + x^(-4) = 7, what is the value of x^8 + x^(-8) =?

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If x^4 + x^(-4) = 7, what is the value of x^8 + x^(-8) =?

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