Find a and b from given questions
Answers
Step-by-step explanation:
Solution :-
i)
Given that (3+√7)/(3-√7) = a+b√7
The denominator = 3-√7
The rationalising factor of 3-√7 is 3+√7
On rationalising the denominator then
=> [(3+√7)/(3-√7)]×[(3+√7)/(3+√7)] = a+b√7
=> [(3+√7)(3+√7)]/[(3-√7)(3+√7)] = a+b√7
=> (3+√7)²/[3²-(√7)²] = a+b√7
Since , (a+b)(a-b) = a²-b²
Where, a = 3 and b = √7
=> (3+√7)²/(9-7) = a+b√7
=> (3+√7)²/2 = a+b√7
=> [(3²+2(3)(√7)+(√7)²]/2 = a+b√7
Since, (a+b)² = a²+2ab+b²
Where, a = 3 and b = √7
=> (9+6√7+7)/2 = a+b√7
=> (16+6√7)/2 = a+b√7
=> 2(8+3√7)/2 = a+b√7
=> 8+3√7 = a+b√7
On comparing both sides then
a = 8 and b = 3
ii)
Given that (5+2√3)/(7+4√3) = a+b√3
The denominator = 7+4√3
The rationalising factor of 7+4√3 is 7-4√3
On rationalising the denominator then
=> [(5+2√3)/(7+4√3)]×[(7-4√3)/(7-4√3)] = a+b√3
=> [(5+2√3)(7-4√3)]/[(7+4√3)(7+4√3)] = a+b√3
=> [(5+2√3)(7-4√3)]/[7²-(4√3)²] = a+b√3
Since , (a+b)(a-b) = a²-b²
Where, a = 7 and b = 4√3
=> [(5+2√3)(7-4√3)]/(49-48) = a+b√3
=> [(5+2√3)(7-4√3)]/1= a+b√3
=> (5+2√3)(7-4√3)= a+b√3
=> (5×7)+(5×-4√3)+(2√3×7)+(2√3×-4√3) = a+b√3
=> 35-20√3+14√3-24 = a+b√3
=> (35-24)+(-20√3+14√3) = a+b√3
=> 11-6√3 = a+b√3
=> 11+(-6)√3 = a+b√3
On comparing both sides then
a = 11 and b = -6
Used formulae:-
→ (a+b)(a-b) = a²-b²
→ (a+b)² = a²+2ab+b²
→ The Rationalising factor of a+√b is a-√b
→ The Rationalising factor of a-√b is a+√b
Answer:
i) a=8 and b=3
ii) a=11 and b=-6
Step-by-step explanation:
Question:
3+√7 / 3-√7 = a+b√3
5+2√3 / 7+4√3 = a+b√3
To find:
Values of a and b in the above two questions.
Solution:
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