if a =√10+√5/√10+√5,and b =√10-√5/√10+√5 then show that √a-√b-2√ab=0 ans me
Answers
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Step-by-step explanation:
Given :-
a =√10+√5/√10-√5
b =√10-√5/√10+√5
To find :-
Show that √a-√b-2√ab=0
Solution:-
Given that:-
a =(√10+√5)/(√10-√5)
Denominator = √10-√5
The Rationalising factor of √10-√5 is √10+√5
On Rationalising the denominator then
a = [(√10+√5)/(√10-√5)] ×[(√10+√5)/(√10+√5)]
=> a = [(√10+√5)(√10+√5)]×[(√10-√5)(√10+√5)]
=> a = (√10+√5)²/[(√10-√5)(√10+√5)]
=> a = (√10+√5)²/[(√10)²-(√5)²]
Since (a+b)(a-b)=a²-b²,
Where a = √10 and b=√5
=> a = (√10+√5)²/(10-5)
=> a = (√10+√5)²/5
=> a = (√10+√5)²/(√5)²
=> a = [(√10+√5)/√5]²
=> √a = √[[(√10+√5)/√5]²]
=> √a = (√10+√5)/√5 -------------(1)
and
b =(√10-√5)/(√10+√5)
Denominator = √10+√5
The Rationalising factor of √10+√5 is √10-√5
On Rationalising the denominator then
=>b = [(√10-√5)/(√10+√5)] ×[(√10-√5)/(√10-√5)]
=> b = [(√10-√5)(√10-√5)]×[(√10+√5)(√10-√5)]
=> b = (√10-√5)²/[(√10+√5)(√10-√5)]
=> b = (√10-√5)²/[(√10)²-(√5)²]
Since (a+b)(a-b)=a²-b²,
Where a = √10 and b=√5
=> b = (√10-√5)²/(10-5)
=> b = (√10-√5)²/5
=> b = (√10-√5)²/(√5)²
=> b = [(√10-√5)/√5]²
=> √b = √[[(√10-√5)/√5]²]
=> √b = (√10-√5)/√5 -------------(2)
On Subtracting (2) from (1) then
√a - √b =
=> [ (√10+√5)/√5]-[(√10-√5)/√5]
=> [(√10+√5)-(√10-√5)]/√5
=> (√10+√5-√10+√5)/√5
=> (√5+√5)/√5
=> 2√5/√5
=> 2
So, √a - √b = 2 -------------------(3)
On multiplying both (1)&(2) then
√a × √b = [ (√10+√5)/√5]× [(√10-√5)/√5]
=>√(ab) = (√10+√5)(√10-√5)/(√5×√5)
=> √(ab) = [(√10)²-(√5)²]/(√5)²
Since (a+b)(a-b)=a²-b²,Where a = √10 and b=√5
=> √(ab) = (10-5)/5
=> √ab = 5/5
=> √ab = 1
On multiplying with 2 both sides then
=> 2√ab = 2-----------------(4)
On subtracting (3) from (4) then
√a -√b - 2√(ab) = 2-2
√a -√b - 2√(ab) = 0
Hence, Proved.
Used formulae:-
- (a+b)(a-b)=a²-b²
- The Rationalising factor of√a+√b is √a-√b
- The Rationalising factor of√a-√b is √a+√b
- (a/b)^m = a^m / a^n