Question

if a2+b2=169,ab=60, (a>b) then(a2-b2)​

Answer 1

Answer:

given that,

a²+b²=169

ab=60

now,

(a+b)²=a²+b²+2ab

(a+b)²=169+2×60

(a+b)²=169+120

(a+b)²=289

rooting both sides

a+b=√289

a+b=17

and,

(a-b)²=a²+b²-2ab

(a-b)²=169-2×60

(a-b)²=169-120

(a-b)²=49

rooting both sides

a-b=√49

a-b=7

now

a²-b²=(a+b)(a-b)

=(17)×(7)

=119

Answer 2

(a+b)² = a²+2ab+b²

But a²+b² = 169 and ab = 60

Therefore (a+b)² = 289

(a+b) = ±17

a+b = 17   or   a+b = -17

Similarly (a-b)² = a²+b²-2ab

(a-b)² = 49

(a-b) = ±7

a-b = 7   or   a-b = -7

a-b = 7 is only possible as a>b

(a+b)(a-b) = a²-b²

Case - I

when a+b = 17

(a+b)(a-b) = a²-b²

17*7 = 119

i.e., a²-b² = 119

Case - II

when a+b = -17

(a+b)(a-b) = a²-b²

-17*7 = -119

i.e., a²-b² = -119

So a²-b² = ±119

HOPE IT HELPS !!