If a = (√5 + √10) / (√10 - √5) and b = (√10 - √5) / (√10 + √5) then show that √a - √b - 2√ab = 0.
Answers
Step-by-step explanation:
Given :-
a = (√5+√10)/(√10-√5)
b = (√10-√5)/(√10+√5)
To find :-
Show that √a - √b - 2√ab = 0.
Solution :-
Given that :
a = (√5+√10)/(√10-√5)
The denominator = √10-√5
The Rationalising factor of√10-√5 is √10+√5
On Rationalising the denominator
=> a =[(√5+√10)/(√10-√5)]×[(√10+√5)/(√10+√5)]
=> a =[(√5+√10)(√10+√5)]×[(√10-√5)(√10+√5)]
=> a = [(√10+√5)(√10+√5)]/[(√10-√5)(√10+√5)]
=> a =(√10+√5)²/[(√10)²-(√5)²]
Since (x+y)(x-y) = x²-y²
Where x = √10 and y = √5
=> a = (√10+√5)²/(10-5)
=> a = (√10+√5)²/5
=> a= [(√10)²+2(√10)(√5)+(√5)²]/5
Since (x+y)² =x²+2xy+y²
=> a = (10+2√50+5)/5
=> a = (15+2√50)/5
=> a = (15+2×5√2)/5
=> a = (15+10√2)/5
=> a = 5(3+2√2)/5
=> a = 3+2√2 ---------------(1)
On square rooting both sides then
=> √a = √(3+2√2)
=> √a = √[(2+1)+2√(2)(√1)]
=> √a = [(√2)²+(√1)²+2(√2)(√1)]
=> √a = √(√2+1)²
=> √a = √2+1 -------------(2)
And
b = (√10-√5)/(√10+√5)
The denominator = √10+√5
The Rationalising factor of√10+√5 is √10-√5
On Rationalising the denominator
=>b = [(√10-√5)/(√10-√5)]×[(√10+√5)/(√10-√5)]
=> b =[(√10-√5)(√10-√5)]×[(√10-√5)(√10+√5)]
=> b =(√10-√5)²/[(√10)²-(√5)²]
Since (x+y)(x-y) = x²-y²
Where x = √10 and y = √5
=> b = (√10-√5)²/(10-5)
=> b = (√10-√5)²/5
=> b= [(√10)²-2(√10)(√5)+(√5)²]/5
Since (x-y)² = x²-2xy+y²
=> b = (10-2√50+5)/5
=>b = (15-2√50)/5
=> b = (15-2×5√2)/5
=> b = (15-10√2)/5
=> b = 5(3-2√2)/5
=> b = 3-2√2 ---------------(3)
On square rooting both sides then
=> √b = √(3-2√2)
=> √b = √[(2+1)-2√(2)(√1)]
=> √b = [(√2)²+(√1)²-2(√2)(√1)]
=> √b = √(√2-1)²
=> √b = √2-1 -------------(4)
On multiplying (2)&(4)
√a×√b = (√2+1)(√2-1)
=> √(ab) = (√2)²-1²
=> √(ab) = 2-1
=>√(ab) = 1-------------(5)
Now,
√a - √b -2√ab
From (2),(4)&(5)
=> (√2+1)-(√2-1)-2(1)
=> √2+1-√2+1-2
=>(√2-√2)+(1+1-2)
=> (0)+(2-2)
=> 0
√a - √b -2√ab= 0
Hence, Proved.
Answer:-
√a - √b -2√ab= 0 for the given problem
Used formulae:-
- (x+y)² =x²+2xy+y²
- (x-y)² =x²-2xy+y²
- (x+y)(x-y) = x²-y²