If a - b = 3 and a2 + b2 = 29, find the value of ab.
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Answered by
4
Hey!
Given,
a-b = 3 and a^2 + b^2 = 29
ATQ,
Given,
a-b = 3 and a^2 + b^2 = 29
ATQ,
2ab = (a^2 + b^2) - (a - b)^2
= 29 - 9
= 20
ab = 10.
Answered by
4
Heya !!!
This is your answer.....
Given :--
a - b = 3.
So ....
a = b + 3.
Putting a = b+3 in a² + b² = 29....
(b+3)² + b² = 29
b² + 6b + 9 + b² = 29
2b² + 6b - 20 = 0.
b² + 3b - 10 = 0. .........(Simplified)
We have to find two numbers a and b such that a+b = 3 and ab = -10.
By looking at the prime factors of 10.. we get a and b equal to 5 and -2 respectively.
Soo.......
b² + 5b - 2b - 10 = 0.
b (b+5) - 2 (b +5) = 0.
(b-2) (b+5) = 0.
Hence b = 2 and b = -5.
Putting these values of b in a - b = 3.
We get .....
a = 5 and a = -2.
Now
a×b = 5×2 = 10
HOPE IT HELPS
This is your answer.....
Given :--
a - b = 3.
So ....
a = b + 3.
Putting a = b+3 in a² + b² = 29....
(b+3)² + b² = 29
b² + 6b + 9 + b² = 29
2b² + 6b - 20 = 0.
b² + 3b - 10 = 0. .........(Simplified)
We have to find two numbers a and b such that a+b = 3 and ab = -10.
By looking at the prime factors of 10.. we get a and b equal to 5 and -2 respectively.
Soo.......
b² + 5b - 2b - 10 = 0.
b (b+5) - 2 (b +5) = 0.
(b-2) (b+5) = 0.
Hence b = 2 and b = -5.
Putting these values of b in a - b = 3.
We get .....
a = 5 and a = -2.
Now
a×b = 5×2 = 10
HOPE IT HELPS
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