Question

solve it (x-2)(x-3)-a+1/a^2=0​

Answer 1

Answer:

see the pic

Step-by-step explanation:

hope it helps

Answer 2

The solution of equation is $x=2-\frac{1}{a},3+\frac{1}{a}$.

Step-by-step explanation:

Given : Equation $(x-2)(x-3)-a+\frac{1}{a^2}=0$

To find : Solve the equation ?

Solution :

Solving the equation,

$x^2-3x-2x+6-a+\frac{1}{a^2}=0$

$x^2-5x-\frac{6a^2-a-1}{a^2}=0$

$x^2-5x-\frac{6a^2-3a+2a-1}{a^2}=0$

$x^2-5x-\frac{3a(2a-1)+1(2a-1)}{a^2}=0$

$x^2-5x-\frac{(2a-1)(3a+1)}{a^2}=0$

Applying middle term split,

$x^2-(\frac{3a+1}{a})x+(\frac{2a-1}{a})x-\frac{(2a-1)(3a+1)}{a^2}=0$

$x(x-\frac{3a+1}{a})+(\frac{2a-1}{a})(x-\frac{3a+1}{a}=0$

$(x-\frac{3a+1}{a})+(x+\frac{2a-1}{a})=0$

$x=\frac{2a-1}{a},\frac{3a+1}{a}$

$x=2-\frac{1}{a},3+\frac{1}{a}$

Therefore, the solution of equation is $x=2-\frac{1}{a},3+\frac{1}{a}$.

#Learn more

Solve (x+3)/(x-3)+(x+2)/(x-3)=2

https://brainly.in/question/7187267