Question

the roots of a^2x^2-(a^2b^2+1)x+b^2=0​

Answer 1

Given,

• The quadratic equation $a^2x^2 -(a^2b^2+1)x +b^2 = 0$ is given.

To find,

• We have to find the roots of the quadratic equation.

Solution,

The roots of the quadratic equation $a^2x^2 -(a^2b^2+1)x +b^2 = 0$ are.

We can simply find the roots of the quadratic equation $a^2x^2 -(a^2b^2+1)x +b^2 = 0$ by taking the terms common out.

                      $a^2x^2 -a^2b^2x-x +b^2 = 0$    

Taking $a^2x$ common from the first two terms and 1 common from last two terms, we get

                      $a^2x (x-b^2) -1(x-b^2)$  

Taking $(x-b^2)$ common, we get

                        $(x-b^2) (a^2x-1) = 0$

Equating the two polynomials equal to 0, we get

                       $(x-b^2)$ = 0,      $(a^2x-1)$ = 0

                        x = b²                 x = 1/a²

∴ The roots of the given equation are b² and 1/a².

Hence, The roots of the quadratic equation $a^2x^2 -(a^2b^2+1)x +b^2 = 0$ are b² and 1/a².

Answer 2

Given,

The quadratic equation a^2x^2 -(a^2b^2+1)x +b^2 = 0a2x2−(a2b2+1)x+b2=0 is given.

To find,

We have to find the roots of the quadratic equation.

Solution,

The roots of the quadratic equation a^2x^2 -(a^2b^2+1)x +b^2 = 0a2x2−(a2b2+1)x+b2=0 are.

We can simply find the roots of the quadratic equation a^2x^2 -(a^2b^2+1)x +b^2 = 0a2x2−(a2b2+1)x+b2=0 by taking the terms common out.

                      a^2x^2 -a^2b^2x-x +b^2 = 0a2x2−a2b2x−x+b2=0    

Taking a^2xa2x common from the first two terms and 1 common from last two terms, we get

                      a^2x (x-b^2) -1(x-b^2)a2x(x−b2)−1(x−b2)  

Taking (x-b^2)(x−b2) common, we get

                        (x-b^2) (a^2x-1) = 0(x−b2)(a2x−1)=0

Equating the two polynomials equal to 0, we get

                       (x-b^2)(x−b2) = 0,      (a^2x-1)(a2x−1) = 0

                        x = b²                 x = 1/a²

∴ The roots of the given equation are b² and 1/a².

Hence, The roots of the quadratic equation a^2x^2 -(a^2b^2+1)x +b^2 = 0a2x2−(a2b2+1)x+b2=0 are b² and 1/a².