Question

Find the zeroes of the polynomial a(x²+1) - x(a²+1) and verify the relationship between the zeroes and the coefficients.​

Answer 1

Answer:

Step-by-step explanation:

a( x² + 1 ) - x( a² + 1 ) = 0 .

==> ax² + a - x( a² + 1 ) = 0 .

==> ax² - ( a² + 1 )x + a = 0 .

Here, A = a , B = -( a² + 1 ) and C = a .

==> ax² - a²x - x + a = 0 .

==> ax( x - a ) - 1( x - a ) = 0 .

==> ( ax - 1 ) ( x - a ) = 0 .

==> ax - 1 = 0 or x - a = 0 .

therefore x = 1/a or a .

Therefore, Sum of zeros = -( coefficient of x )/( coefficient of x² ) = - B/A

==> 1/a + a = -( -( a² + 1 ) )/ a .

therefore ( 1 + a² )/a = ( 1 + a² )/ a .

And, Product of zeros = Constant term/ coefficient of x² = C/A .

==> 1/a × a = a/a .

therefore, 1 = 1 .

Hope it helps!

Answer 2

Answer:

1/a or a .

Step-by-step explanation :

We have,

→ A quadratic polynomial :

→ a( x² + 1 ) - x( a² + 1 ) = 0 .

==> ax² + a - x( a² + 1 ) = 0 .

==> ax² - ( a² + 1 )x + a = 0 .

Here, A = a , B = -( a² + 1 ) and C = a .

==> ax² - a²x - x + a = 0 .

==> ax( x - a ) - 1( x - a ) = 0 .

==> ( ax - 1 ) ( x - a ) = 0 .

==> ax - 1 = 0 or x - a = 0 .

•°• x = 1/a or a .

VERIFICATION :)

Therefore, Sum of zeros = -( coefficient of x )/( coefficient of x² ) = - B/A

==> 1/a + a = -( -( a² + 1 ) )/ a .

•°• ( 1 + a² )/a = ( 1 + a² )/ a .

And, Product of zeros = Constant term/ coefficient of x² = C/A .

==> 1/a × a = a/a .

•°• 1 = 1 .

hope it helps you ❤️✨