Q. 21 B4.9( 14 Votes )

Solve the followi

Answer :

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,

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