Question

find the roots of a(x²+1)-x(a²+1)​

Answer 1

Answer:

Given, quadratic equation is

=> a(x2 + 1) - x(a2 + 1) = 0

=> ax2 + a - xa2 - x = 0

=> ax2 - (a2 + 1)x + a = 0

Using quadratic formula, we get

=> x = [-{-(a2 + 1)} ± √{(a2 + 1)2 - 4 * a * a}]/2a

=> x = [(a2 + 1) ± √{a4 + 1 + 2a2 - 4 * a2 }]/2a

=> x = [(a2 + 1) ± √{a4 + 1 - 2a2 }]/2a

=> x = [(a2 + 1) ± √{(a2 - 1)2 }]/2a

=> x = [(a2 + 1) ± (a2 - 1)]/2a

=> x = [(a2 + 1) + (a2 - 1)]/2a and x = [(a2 + 1) - (a2 - 1)]/2a

=> x = 2a2 /2a and x = 2/2a

=> x = a, 1/a

So, the roots are a and 1/a

Answer 2

Answer:

a(x^2+1)-x(a^2+1)=

a*x^2 + a - (a^2)*x - x=

a*x^2 - (a^2+1)*x + a=

(a*x-1)*(x-a)

When a*x-1=0, x=(1/a)

When x-a=0, x = a.

So the roots are (1/a) and a.