Math, asked by kaushalkarnakayansh, 1 month ago

If p=9-(79)^0.5, Find value of p + 2/p

Answers

Answered by tennetiraj86
3

Step-by-step explanation:

Given:-

P = 9-(79)^0.5

To find:-

Find the value of P +(2/P)?

Solution:-

Given that :

P = 9-(79)^0.5

=> P = 9-(79)^(5/10)

=> P = 9-(79)^(1/2)

=> P = 9-√79

Now

1/P

= 1/(9-√79)

The denominator = 9-√79

We know that

The Rationalising factor of a -√b = a+√b

Rationalising factor of 9-√79 = 9+√79

On Rationalising the denominator then

=> 1(9+√79)/[(9-√79)(9+√79)]

=> (9+√79)/[(9-√79)(9+√79)]

We know that

(a+b)(a-b)=a^2-b^2

Where , a= 9 and b = √79

=> (9+√79)/[9^2-(√79)^2]

=> (9+√79)/(81-79)

=> (9+√79)/2

1/P = (9+√79)/2

2/P = 2[(9+√79)/2]

2/P = 9+√79

Now the value of P+(1/P)

=> (9-√79) + (9+√79)

=> 9-√79+9+√79

=> 9+9

=> 18

P+(2/P)=18

Answer:-

The value of P+(2/P) for the given problem is 18

Used formulae:-

  • The Rationalising factor of a -√b = a+√b
  • (a+b)(a-b)=a^2-b^2

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