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Expansion of (x-y)^2

If a-1/a=4, t...

Question

If $a - \frac{1}{a}$ = 4 , then $a^{2} + \frac{1}{a^{2}}$ = 16

A

True

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B

False

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Solution

The correct option is B
False

Given that
$a - \frac{1}{a}$ = 4
$s q u a r i n g o n b o t h s i d e s$
$(a - \frac{1}{a})^{2}$ = 16
$a^{2} + \frac{1}{a^{2}} - 2$ = 16
$a^{2} + \frac{1}{a^{2}}$ = 18

Hence the given statement is false

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Similar questions

a - \frac{1}{a} = 4

a^{2} + \frac{1}{a^{2}}

a + \frac{1}{a} = 4

a \neq 0

a^{2} + \frac{1}{a^{2}}

.

a + \frac{1}{a} = 4

, then the value of

a^{2} - \frac{1}{a^{2}}

a \neq 0

a - \frac{1}{a} = 4

a^{2} + \frac{1}{a^{2}}

a^{2} - 4 a + 1 = 4

, then the value of

\frac{a^{3} - a^{2} + a - 1}{a^{2} - 1} (a^{2} \neq 1)

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