Question

if a-1/a=3 then find a2+1/a2​

Answer 1

$\huge\mathbb{SOLUTION:-}$

$\sf\bullet a-\dfrac{1}{a}=3$

• We have to find the value of

$\sf\bullet a{}^{2}+\dfrac{1}{a{}^{2}}=?$

Now formula used

$\boxed{\sf{(a-b){}^{2}=a{}^{2}+b{}^{2}-2ab}}$

So

$\sf \bigg(a-\dfrac{1}{a}\bigg){}^{2}=a{}^{2}+\dfrac{1}{a{}^{2}}-2\times \cancel{a}\times \dfrac{1}{\cancel {a}}\\ \\ \sf\implies a{}^{2}+\dfrac{1}{a{}^{2}}-2$

$\sf\underline{\red{A.T.Q:-}}$

$\sf\implies \bigg(a-\dfrac{1}{a}\bigg){}^{2}=a{}^{2}+\dfrac{1}{a{}^{2}}-2\\ \\ \sf\implies (3){}^{2}=a{}^{2}+\dfrac{1}{a{}^{2}}-2\\ \\ \sf\implies 9-2=a{}^{2}+\dfrac{1}{a{}^{2}}\\ \\ \sf\implies a{}^{2}+\dfrac{1}{a{}^{2}}=6$

$\boxed{\purple{\sf{a{}^{2}+\dfrac{1}{a{}^{2}}=6}}}$

Answer 2

$\huge\bold\green{Question}$

If a-1/a=3 then find a²+1/a²

$\huge\bold\green{Answer}$

According to the question we have to find the

value of a²+1/a²

Simply, by using the identity

$\tt\green{(a-b){}^{2}=a{}^{2}+b{}^{2}-2ab}$

$\begin{lgathered}\tt \bigg(a-\dfrac{1}{a}\bigg){}^{2}=a{}^{2}+\dfrac{1}{a{}^{2}}-2\times \cancel{a}\times \dfrac{1}{\cancel {a}}\\ \\ \tt = a{}^{2}+\dfrac{1}{a{}^{2}}-2\end{lgathered}$

So, according to the question,

$\begin{lgathered}\tt \bigg(a-\dfrac{1}{a}\bigg){}^{2}=a{}^{2}+\dfrac{1}{a{}^{2}}-2\\ \\ \tt = (3){}^{2}=a{}^{2}+\dfrac{1}{a{}^{2}}-2\\ \\ \tt = 9-2=a{}^{2}+\dfrac{1}{a{}^{2}}\\ \\ \tt = a{}^{2}+\dfrac{1}{a{}^{2}}=6\end{lgathered}$

Hence, the required answer is $\tt{a{}^{2}+\dfrac{1}{a{}^{2}}=6}$