If $$a^2+b^2=50$$​ and a×b =7. Find $$\frac{(a-b)}{(a+b)}$$​, where a>b.

Correct option is D

Given:

​$$a^2+b^2=50$$

ab = 7

Concep used:

Using the identity for $$(a + b)^2$$​ and $$(a - b)^2$$

​$$(a+b)^2=a^2+b^2+2ab$$

​​$$(a−b)^2=a^2+b^2−2ab$$​​

Solution:

we calculate a + b and a - b:

​$$(a+b)^2$$ = 50 +2 $$\times$$​7 = 50+14 = 64

​$$(a−b)^2$$ = 50- 2$$\times$$​7 = 50-14 = 36

So,

a+b = $$\sqrt64$$ = 8 (take positive root as a +b > 0)

​a−b = $$\sqrt36$$ = 6 (take positive root as a>b)

​Now, we can find $$\frac{a−b}{a+b}:$$

​​$$\frac{a−b}{a+b}$$ = $$\frac{3}4$$

Thus, correct option is (d)