JAMB - Mathematics (1990 - No. 17)
If x + \(\frac{1}{x}\) = 4, find x2 + \(\frac{1}{x^2}\)
16
14
12
9
Тайлбар
x + \(\frac{1}{x}\) = 4, find x2 + \(\frac{1}{x^2}\)
= (x + \(\frac{1}{x}\))2 = x2 + \(\frac{1}{x^2}\) + 2
x2 + \(\frac{1}{x^2}\) = ( x + \(\frac{1}{x^2}\))2 - 2
= (4)2 - 2
= 16 - 2
= 14
= (x + \(\frac{1}{x}\))2 = x2 + \(\frac{1}{x^2}\) + 2
x2 + \(\frac{1}{x^2}\) = ( x + \(\frac{1}{x^2}\))2 - 2
= (4)2 - 2
= 16 - 2
= 14