What is t if t^8=8^2 ?

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to solve for t given that t^8 = 8^2

t^8 = 8^2

=> t^8 = 2^3^2

=> t^8 = 2^6

=> t = 2^(6/8)

=> t = 2^(3/4)

The required value of t = 2^(3/4)

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

Since we need the formula of difference of squares to solve the equation, we'll recall first the formula for the difference of squares:

a^2 - b^2 =(a-b)(a+b)

Now,we'll put a^2 = t^8 = (t^4)^2 and b^2 = 64 = 8^2

We'll re-write the given equation, emphasizing on the difference of squares:

 (t^4)^2 - 8^2 = (t^4 - 8)(t^4 + 8)

t^4 - 8 is also a difference o squares, whose terms are: a = t^2 and b = 2sqrt2

(t^4 - 8)(t^4 + 8) = (t^2 - 2sqrt2)(t^2 + 2sqrt2)(t^4 + 8)

Now, we'll solve the equation:

t^8 - 8^2 = 0

t^8 - 8^2 = (t^2 - 2sqrt2)(t^2 + 2sqrt2)(t^4 + 8)

(t^2 - 2sqrt2)(t^2 + 2sqrt2)(t^4 + 8) = 0

t^2 - 2sqrt2 = 0

t^2 = 2sqrt2

t1 = +sqrt(2sqrt2)

t2 = -sqrt(2sqrt2)

t^2 + 2sqrt2 = 0

t3 = +i*sqrt(2sqrt2)

t4 = -i*sqrt(2sqrt2)

The roots of the equation, both real and imaginary, are {+sqrt(2sqrt2) ; -sqrt(2sqrt2) ; +i*sqrt(2sqrt2) ; -i*sqrt(2sqrt2)}.

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