Question

$if (a - \frac{1}{a} ) = 6 \: find \: {a}^{2} + \frac{1}{a}^{2} }$
​

Answer 1

Given:

$\sf\hookrightarrow \bigg(a-\dfrac 1 a\bigg) =6$

$\rule{240}{1}$

To find:

$\sf \hookrightarrow a^2+\bigg(\dfrac 1 a\bigg )^2$

$\rule{240}{1}$

Solution:

Squaring both side of given equation

$\hookrightarrow\sf\bigg( a -\dfrac 1 a\bigg)^2=(6)^2$

Applying algebraic identity (a-b)²=a²+b²-2ab.

$\hookrightarrow \sf (a)^2+\bigg( \dfrac 1 a\bigg)^2-2(a)\bigg(\dfrac 1 a\bigg)=(6)^2$

Now solving it

$\hookrightarrow \sf a^2+ \dfrac 1 {a^2 }-2(\not a)\bigg(\dfrac 1 {\not a}\bigg)=36$

$\hookrightarrow \sf a^2+ \dfrac 1 {a^2 }-2=36$

$\hookrightarrow \sf a^2+ \dfrac 1 {a^2 }=36+2$

$\hookrightarrow \sf a^2+ \dfrac 1 {a^2 }=38$

$\boxed{\blue{\hookrightarrow \sf a^2+ \bigg( \dfrac 1 {a }\bigg)^2=38}}$

So the required answer is 38.

Answer 2

Answer:

Given:

\sf\hookrightarrow \bigg(a-\dfrac 1 a\bigg) =6↪(a−

a

1

)=6

\rule{240}{1}

To find:

\sf \hookrightarrow a^2+\bigg(\dfrac 1 a\bigg )^2↪a

2

+(

a

1

)

2

\rule{240}{1}

Solution:

Squaring both side of given equation

\hookrightarrow\sf\bigg( a -\dfrac 1 a\bigg)^2=(6)^2↪(a−

a

1

)

2

=(6)

2

Applying algebraic identity (a-b)²=a²+b²-2ab.

\hookrightarrow \sf (a)^2+\bigg( \dfrac 1 a\bigg)^2-2(a)\bigg(\dfrac 1 a\bigg)=(6)^2↪(a)

2

+(

a

1

)

2

−2(a)(

a

1

)=(6)

2

Now solving it

\hookrightarrow \sf a^2+ \dfrac 1 {a^2 }-2(\not a)\bigg(\dfrac 1 {\not a}\bigg)=36↪a

2

+

a

2

1

−2(



a)(



a

1

)=36

\hookrightarrow \sf a^2+ \dfrac 1 {a^2 }-2=36↪a

2

+

a

2

1

−2=36

\hookrightarrow \sf a^2+ \dfrac 1 {a^2 }=36+2↪a

2

+

a

2

1

=36+2

\hookrightarrow \sf a^2+ \dfrac 1 {a^2 }=38↪a

2

+

a

2

1

=38

\boxed{\blue{\hookrightarrow \sf a^2+ \bigg( \dfrac 1 {a }\bigg)^2=38}}

↪a

2

+(

a

1

)

2

=38

So the required answer is 38.