Solve the followi

When we compare the above quadratic equation with the generalized one we get,

ax² + bx + c = 0

a = 9

b = – 9(a + b)

c = (2a² + 5ab + 2b²)

There is one formula developed by Shridharacharya to determine the roots of a quadratic equation which is as follows:

Before putting the values in the formula let us check the nature of roots by b² – 4ac >0

⟹ (– 9(a + b)) ² – (4 × 9 × (2a² + 5ab + 2b²))

⟹ 81(a² +2ab + b²) - (36(2a² + 5ab + 2b²))

⟹ 81a² +162ab + 81b² - 72a² - 180ab - 72b²

⟹ 9a² - 18ab + 9b²

⟹ 9(a² - 2ab + b²)

⟹ 3²(a – b) ²

Now let us put the values in the above formula

Solving with positive value first,

Solving with negative value second,