If $$a - \frac{1}{a} = 7,$$ then $$a^2 + \frac{1}{a^2} = ?$$
we know that:
$$(a-b)^2 = a^2 + b^2 -2ab$$
and
$$(a+b)^2 = a^2 + b^2 +2ab$$
here,
$$(a-\frac{1}{a}) = 7$$
Squaring on both side .....
$$(\frac{a-1}{a})^2 = 7² = 49$$
now,
$$(\frac{a+1}{a})^2 = (\frac{a-1}{a})^2 + 4 \times a\times \frac{1}{a}$$
$$(\frac{a+1}{a})^2 = 49 + 4$$
$$(\frac{a+1}{a})^2 = 53$$
$$a^2 + \frac{1}{a^2} +2a\frac{1}{a} =53$$
$$a^2 + \frac{1}{a^2} = 53-2 = 51$$
Option B is correct.
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