Question
If $a+\frac{1}{a}=6$ and $a \neq 0$ find:$(i) a-\frac{1}{a};(ii) a^2-\frac{1}{a^2}$
✓
Answer
We know that,
$(a+b)^2=a^2+2 a b+b^2$
and
$(a-b)^2=a^2-2 a b+b^2$
Thus,
$\left(a+\frac{1}{a}\right)^2=a^2+\frac{1}{a^2}+2 \times a \times \frac{1}{a}$
$=a^2+\frac{1}{a^2}+2 \dots.....(1)$
Given that $a+\frac{1}{a}=6$; Substitute in equation $(1),$ we have
$(6)^2=a^2+\frac{1}{a^2}+2$
$\Rightarrow a^2+\frac{1}{a^2}=36-2$
$\Rightarrow a^2+\frac{1}{a^2}=34\ldots...(2)$
Similarly, consider
$\left(a-\frac{1}{a}\right)^2=a^2+\frac{1}{a^2}-2 \times a \times \frac{1}{a}$
$=a^2+\frac{1}{a^2}-2$
$=34-2 [$ from $(2) ]$
$ \Rightarrow\left(a-\frac{1}{a}\right)^2=32$
$ \Rightarrow\left(a-\frac{1}{a}\right)= \pm \sqrt{32}$
$\Rightarrow\left(a-\frac{1}{a}\right)= \pm 4 \sqrt{2}\ldots....(3)$
$(ii)$ We need to find $a^2-\frac{1}{a^2}$
We know that, $a^2-\frac{1}{a^2}=\left(a-\frac{1}{a}\right)\left(a+\frac{1}{a}\right)$
$a-\frac{1}{a}= \pm 4 \sqrt{2} ; a+\frac{1}{a}=6$
Thus,
$a^2-\frac{1}{a^2}=( \pm 4 \sqrt{2})(6)$
$ \Rightarrow a^2-\frac{1}{a^2}=( \pm 24 \sqrt{2})$
$(a+b)^2=a^2+2 a b+b^2$
and
$(a-b)^2=a^2-2 a b+b^2$
Thus,
$\left(a+\frac{1}{a}\right)^2=a^2+\frac{1}{a^2}+2 \times a \times \frac{1}{a}$
$=a^2+\frac{1}{a^2}+2 \dots.....(1)$
Given that $a+\frac{1}{a}=6$; Substitute in equation $(1),$ we have
$(6)^2=a^2+\frac{1}{a^2}+2$
$\Rightarrow a^2+\frac{1}{a^2}=36-2$
$\Rightarrow a^2+\frac{1}{a^2}=34\ldots...(2)$
Similarly, consider
$\left(a-\frac{1}{a}\right)^2=a^2+\frac{1}{a^2}-2 \times a \times \frac{1}{a}$
$=a^2+\frac{1}{a^2}-2$
$=34-2 [$ from $(2) ]$
$ \Rightarrow\left(a-\frac{1}{a}\right)^2=32$
$ \Rightarrow\left(a-\frac{1}{a}\right)= \pm \sqrt{32}$
$\Rightarrow\left(a-\frac{1}{a}\right)= \pm 4 \sqrt{2}\ldots....(3)$
$(ii)$ We need to find $a^2-\frac{1}{a^2}$
We know that, $a^2-\frac{1}{a^2}=\left(a-\frac{1}{a}\right)\left(a+\frac{1}{a}\right)$
$a-\frac{1}{a}= \pm 4 \sqrt{2} ; a+\frac{1}{a}=6$
Thus,
$a^2-\frac{1}{a^2}=( \pm 4 \sqrt{2})(6)$
$ \Rightarrow a^2-\frac{1}{a^2}=( \pm 24 \sqrt{2})$
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