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