Question 62

If $$a - \frac{1}{a} = 7,$$ then $$a^2 + \frac{1}{a^2} = ?$$

Solution

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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