Q. 21 A5.0( 1 Vote )

# Solve the followi

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

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :

If u(x) = (x – 1)Rajasthan Board Mathematics

Find the lowest cRajasthan Board Mathematics

Solve the followiRajasthan Board Mathematics

Solve the followiRajasthan Board Mathematics

The area of a recRajasthan Board Mathematics

The LCM and HCF oRajasthan Board Mathematics

If α and β are thRajasthan Board Mathematics

Find the zeroes oRajasthan Board Mathematics

Write ShridharachRajasthan Board Mathematics

Solve the followiRajasthan Board Mathematics