Question

If a-b=5 and a2+b2=37,find the value of ab

Answer 1

Given that,

(a-b) = 5

a = 5+b ...eq1

a^2 + b^2 = 37 ...eq2

Now, using eq1 and eq2, we get

(5+b)^2 + b^2 = 37

》Consider the identity

(a+b)^2 = a^2+b^2+2ab

Applying the identity

5^2 +b^2 +2×5b +b^2 = 37

25 + b^2 + 10b +b^2 = 37

10b +2b^2 = 37 -25

2(5b+b^2) = 12

5b+b^2 = 12/2 = 6

b^2 +5b -6 = 0

》By Quadratic Formula

x = -b+-√(b^2-4ac) /2a

Applying the formula , we get

b = -5+-√25-{4×1×(-6)} /2×1

b = -5+- √25+24 / 2

b = -5+- √49 / 2

b = -5+- 7 / 2

Now,

b = -5+7/2

b = 2/2 = 1

or

b = -5-7 /2

b = -12 /2

b = -6

Similarly,

(a-b) = 5

b = a-5 ...eq3

(a^2 +b^2) = 37 ...eq4

Now, using eq3 and eq4

a^2 +(a-5)^2 =37

》Consider the identity

(a-b)^2 = a^2+b^2-2ab

a^2 +a^2 +25 -2a(-5) = 37

2a^2 +10a = 37 -25

2(a^2+5a) = 12

a^2 +5a = 12/2

a^2 +5a-6 = 0

》By Factorization method

a^2 +5a -6 = 0

a^2 +6a -a -6 = 0

a(a+6) -1(a+6) = 0

(a-1) (a+6) =0

Now,

a-1 = 0

a = 1

or

a+6 = 0

a = -6

Here,

a-b = 5

So, a = -6

and b = 1

Thus

a×b = -6×1

= -6

Hope it helps !