Question

Match the following with the value of \(a^2+\frac{1}{a^2}\) for the conditions given below.

Answer 1

Using the identity,
\((a+b)^2 =a^2 + 2ab +b^2 \),
\(a^2+b^2 =(a+b)^2 - 2ab \).

So, \(a^2+\frac{1}{a^2} = (a+\frac{1}{a})^2-2\)

(i) Given that \(a+\frac{1}{a}=4\)
\(\implies a^2+\frac{1}{a^2}=(a+\frac{1}{a})^2-2 = 4^2 - 2 \)
\(a^2+\frac{1}{a^2} = 16 - 2 = 14\)

(ii) Given that \(a+\frac{1}{a}=6\)
\(\implies a^2+\frac{1}{a^2}=(a+\frac{1}{a})^2-2 = 6^2 - 2 \)
\(a^2+\frac{1}{a^2} = 36 - 2 = 34\)

​Using identity,
​​​​​​\((a-b)^2 =a^2 - 2ab +b^2 \),
\(a^2+b^2=(a-b)^2+2ab\).
So,​​​​​ \(a^2+\frac{1}{a^2} = (a-\frac{1}{a})^2 + 2 \)

(iii) Given that \(a-\frac{1}{a}=5\)
\(\implies a^2+\frac{1}{a^2}= (a-\frac{1}{a})^2 + 2 = 5^2 + 2= 27\)

(iv) Given that \(a-\frac{1}{a}=3\)
\(\implies a^2+\frac{1}{a^2}= (a-\frac{1}{a})^2 + 2 = 3^2 + 2= 11\)