Math, asked by nanditakundu69, 8 months ago

A^2-5a-1=0, find a+1/a,a^2-1/a^2​

Answers

Answered by nishesh2009chaudhary
0

Step-by-step explanation:

a+\frac{1}{a}=\sqrt{29}a+

a

1

=

29

a^2-\frac{1}{a^2}=5\sqrt{29}a

2

a

2

1

=5

29

Step-by-step explanation:

Given : Equation a^2-5a-1=0a

2

−5a−1=0

To find : a+\frac{1}{a},a^2-\frac{1}{a^2}a+

a

1

,a

2

a

2

1

Solution :

Re-write the equation as,

a^2-1=5aa

2

−1=5a

a(a-\frac{1}{a})=5aa(a−

a

1

)=5a

a-\frac{1}{a}=5a−

a

1

=5

Taking, (a+\frac{1}{a})^2=(a-\frac{1}{a})^2+4a\times \frac{1}{a}(a+

a

1

)

2

=(a−

a

1

)

2

+4a×

a

1

(a+\frac{1}{a})^2=(5)^2+4(a+

a

1

)

2

=(5)

2

+4

(a+\frac{1}{a})^2=25+4(a+

a

1

)

2

=25+4

(a+\frac{1}{a})^2=29(a+

a

1

)

2

=29

Taking root both side,

a+\frac{1}{a}=\sqrt{29}a+

a

1

=

29

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

2

a

2

1

=(a+

a

1

)(a−

a

1

)

a^2-\frac{1}{a^2}=(\sqrt{29})(5)a

2

a

2

1

=(

29

)(5)

a^2-\frac{1}{a^2}=5\sqrt{29}a

2

a

2

1

=5

29

#Learn more

Simplify:√(a^2-b^2)+a/√(a^2+b^2)+b ÷√(a^2+b^2)-b/a-√(a^2-b^2)

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