Question

Find the quadratic equation whose roots are its own reciprocal

Answer 1

let roots be p and q
so quadratic equation = ax^2 + bx+c=0
for the required quadratic equation outside are 1/p and =1/q
we know that sum of roots =-b/a and product of roots=c/a
1/p+1/q=-b/a
q+p/xy=-b/a-----(1)

1/p× 1/q = c/a
1/pq=c/a-----(2)
from 1 and 2
a=pq
b= -(p+q)
c=1
required quadratic equation:
pqx^2 -(p+q)x +1

Answer 2

Let one root of the quadratic equation be n, then the other root will be (1/n)

Sum of the roots
= (n) + (1/n)
= (n^2+1)/n

Product of the roots
= (n)×(1/n)
= 1

Quadratic Equation ;
x^2 - (sum of roots) x + (product of roots) = 0
x^2 -[(n^2+1)/n]x + 1 = 0

multiply both sides by n :

nx^2 -n^2x + x + nx = 0
nx(x-n+1) + x = 0
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