Question

If a^2-4a-1=0 then find the value of a^2+1÷a^2

Answer 1

Question:

If a^2 - 4a - 1 = 0 , then find the value of

a^2 + 1/a^2 .

Answer:

a^2 + 1/a^2 = 18

Note:

• (A+B)^2 = A^2 + B^2 + 2•A•B

• (A-B)^2 = A^2 + B^2 - 2•A•B

• A^2 - B^2 = (A+B)(A-B)

Solution:

We have;

=> a^2 - 4a - 1 = 0

=> a^2 - 1 = 4a

=> a - 1/a = 4 ------------(1)

Squaring both sides of eq-(1) , we get;

=> (a - 1/a)^2 = 4^2

=> a^2 + 1/a^2 - 2•a•(1/a) = 16

=> a^2 + 1/a^2 - 2 = 16

=> a^2 + 1/a^2 = 16 + 2

=> a^2 + 1/a^2 = 18

Hence,

The required value of a^2 + 1/a^2 is 18.

Answer 2

Answer:-

$\implies \: \pink{ \boxed{{\green{\boxed{\red{\bf{ {a}^{2} + \frac{1}{ {a}^{2} } = 18}}}}}}} \:$

Step - by - step explanation:-

Used identity :-

$\star \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy$

Solution :-

According to the question,

$\implies \: {a}^{2} - 4a - 1 = 0 \\ \\ \implies \: {a}^{2} - 1 = 4a \\ \\ \: \pink{\bf{divided \: by \: \: a \: \: on \: both \: sides}} \: \\ \\ \implies \: \frac{ {a}^{2} }{a} - \frac{1}{a} = \frac{4a}{a} \\ \\ \implies \: a - \frac{1}{a} = 4 \\ \\ \red{\bf{now \: squaring \: on \: both \: sides}} \\ \\ \implies \: { \bigg(a - \frac{1}{a} \bigg)} ^{2} = {(4)}^{2} \\ \\ \green{ \bf{using \: the \: given \: identity}} \\ \\ \implies {a}^{2} + { \bigg(\frac{1}{a} \bigg)}^{2} - 2 \times a \times \frac{1}{a} = 16 \\ \\ \implies \: {a}^{2} + \frac{1}{ {a}^{2} } - 2 = 16 \\ \\ \implies \: {a}^{2} + \frac{1}{ {a}^{2} } = 16 + 2 \\ \\ \implies \: \pink{ \boxed{{\green{\boxed{\red{\bf{ {a}^{2} + \frac{1}{ {a}^{2} } = 18}}}}}}}$

Hope it helps you.