Question

if a square -5a +1 =and ais not equal to 0 then find a-1/a and a+1/a​

Answer 1

Given:-

a² - 5a + 1 = 0 (a≠0) ....[1]

To find:-

$\rm a-\dfrac{1}{a} \ and \ a+\dfrac{1}{a}$

Solution:-

Dividing [1] by a

$\rm\dashrightarrow\dfrac{a^2-5a+1=0}{a}$

$\rm\dashrightarrow a-5+\dfrac{1}{a} = 0$

Or

$\bf\dashrightarrow a+\dfrac{1}{a}=5$

Squaring both the sides,

$\rm\dashrightarrow \Bigg ( a+\dfrac{1}{a} \Bigg )^2=(5)^2$

Since, (a+b)² = a²+b²+2ab, Hence,

$\rm\dashrightarrow a^2+\dfrac{1}{a^2}+2\times a\times \dfrac{1}{a}=25$

$\rm\dashrightarrow a^2+\dfrac{1}{a^2}+2=25$

$\rm\dashrightarrow a^2+\dfrac{1}{a^2}=25-2$

$\rm\dashrightarrow a^2+\dfrac{1}{a^2}=23$

Now we will find $\rm a-\dfrac{1}{a}$

We know that,

(a-b)² = a²+b²-2ab

Therefore,

$\rm\dashrightarrow \Bigg ( a-\dfrac{1}{a} \Bigg )^2 = a^2 + \dfrac{1}{a^2}-2\times a\times \dfrac{1}{a}$

$\rm\dashrightarrow \Bigg ( a-\dfrac{1}{a} \Bigg )^2 = a^2 + \dfrac{1}{a^2}-2$

$\rm\dashrightarrow \Bigg ( a-\dfrac{1}{a} \Bigg )^2 = 23-2$

$\rm\dashrightarrow \Bigg ( a-\dfrac{1}{a} \Bigg )^2 = 21$

$\bf\dashrightarrow a-\dfrac{1}{a} = \sqrt{21}$