Q. 21 A5.0( 1 Vote )

Solve the followi

Answer :

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

ax2 + bx + c = 0


a = p2


b = (p2 – q2)


c = – q2


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


((p2 – q2)) 2 – (4 × p × – q2)


(p4 –2 p2 q2 + q4) - (-4p2q2)


p4 –2 p2 q2 + q4 + 4p2q2


(p4 + 2 p2 q2 + q4)


((p2 + q2)) 2


Now let us put the values in the above formula




Solving with positive value first,




x = q2 / p2


Solving with negative value second,




x = -1


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