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jmc

algebra intermediate

Problem

Given that

x^{2} + \frac{1}{x ^{2}} = 7

, what is the value of

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

?

Solution

Observe that

(x^{2} + \frac{1}{x ^{2}})^{2} = x^{4} + 2 \cdot x^{2} (\frac{1}{x ^{2}}) + \frac{1}{x ^{4}} = x^{4} + \frac{1}{x ^{4}} + 2.

Therefore,

x^{4} + \frac{1}{x ^{4}} = (x^{2} + \frac{1}{x ^{2}})^{2} - 2 = 7^{2} - 2 = 47

.

Final answer

47