Question

A+1/a=6 .find the value of a^4+1/a^4​

Answer 1

Question:

If a + 1/a = 6 , then find the value of :

a^4 + 1/a^4.

Note:

(x+y)^2 = x^2 + y^2 + 2•x•y

Solution:

We have;

a + 1/a = 6 ---------(1)

We know that;

(x+y)^2 = x^2 + y^2 + 2•x•y

Thus;

=> (a +1/a)^2 = a^2 + 1/a^2 + 2•a•(1/a)

=> (6)^2 = a^2 + 1/a^2 + 2

{ using eq-(1) }

=> 36 = a^2 + 1/a^2 + 2

=> a^2 + 1/a^2 = 36 - 2

=> a^2 + 1/a^2 = 34 -----------(2)

Again;

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

+ 2•(a^2)•(1/a^2)

=> (34)^2 = a^4 + 1/a^4 + 2

{ using eq-(2) }

=> 1156 = a^4 + 1/a^4 + 2

=> a^4 + 1/a^4 = 1156 - 2

=> a^4 + 1/a^4 = 1154

Hence,

The required value of x^4 + 1/x^4 is;

1154.

Answer 2

Answer:-

$\implies \: \boxed{ \bf{ {a}^{4} + \frac{1}{ {a}^{4} } = 1154}}$

Step - by - step explanation:-

To find :-

$\implies \: {a}^{4} + \frac{1}{ {a}^{4} } =$

Formula used :-

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

Solution :-

According to the question→

$\bf{ \rightarrow \: a + \frac{1}{a} = 6} \\ \\ \red{ \bf{squaring \: on \: both \: sides \: }} \\ \\ \rightarrow \: { \bigg(a + \frac{1}{a} \bigg)}^{2} = {(6)}^{2} \\ \\ \bf{using \: the \: given \: formula \: } \\ \\ \: \rightarrow \: \bf{ {a}^{2} + \frac{1}{ {a}^{2} } + 2 \times a \times \frac{1}{a} = 36} \\ \\ \rightarrow \: \bf{ {a}^{2} + \frac{1}{ {a}^{2} } = 36 - 2} \\ \\ \rightarrow \: \bf{ {a}^{2} + \frac{1}{ {a}^{2} } = 34} \\ \\ \green{ \bf{ again \: squaring \: on \: both \: sides}} \\ \\ \rightarrow \: \bf{ { \bigg( {a}^{2} + \frac{1}{ {a}^{2} } \bigg)}^{2} = {(34)}^{2} } \\ \\ \pink{\bf{ again \: using \: the \: given \: formula}}$ $\rightarrow \: \bf{{a}^{4} + \frac{1}{ {a}^{4} } + 2 \times {a}^{2} \times \frac{1}{ {a}^{2} } = 1156} \\ \\ \rightarrow \: \bf{ {a}^{4} + \frac{1}{ {a}^{4}} = 1156 - 2 } \\ \\ \rightarrow \: \boxed{ \bf{ {a}^{4} + \frac{1}{ {a}^{4} } = 1154}}$

Hope it helps you.