Question

a (x² +1) – x(a²+1) = 0​

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

Answer 2

Answer:

ax²+a-ax²-x=0

a-x=0

Step-by-step explanation:

So here one is ax²and another is -ax² and the other pair is a-x.So the variables a and x are same number