Answer :

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

ax^{2} + bx + c = 0

a = p^{2}

b = (p^{2} – q^{2})

c = – q^{2}

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 b^{2} – 4ac >0

⟹ ((p^{2} – q^{2})) ^{2} – (4 × p × – q^{2})

⟹ (p^{4} –2 p^{2} q^{2} + q^{4}) - (-4p^{2}q^{2})

⟹ p^{4} –2 p^{2} q^{2} + q^{4} + 4p^{2}q^{2}

⟹ (p^{4} + 2 p^{2} q^{2} + q^{4})

⟹ ((p^{2} + q^{2})) ^{2}

Now let us put the values in the above formula

Solving with positive value first,

x = q^{2} / p^{2}

Solving with negative value second,

x = -1

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