The given equation is a quadratic equation in the standard form:

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

Steps to solve the quadratic equation:

1. The quadratic equation is of the form: $Ax^2 + Bx + C = 0$ Here:

   • $A = b$
   • $B = -(a+b)$
   • $C = a$

2. Use the quadratic formula: $x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$

3. Substitute $A$, $B$, and $C$ into the formula: $x = \frac{-(-(a+b)) \pm \sqrt{(-(a+b))^2 - 4(b)(a)}}{2(b)}$

4. Simplify step by step: $x = \frac{a+b \pm \sqrt{(a+b)^2 - 4ab}}{2b}$

5. Expand the discriminant $(a+b)^2 - 4ab$: $(a+b)^2 - 4ab = a^2 + 2ab + b^2 - 4ab = a^2 - 2ab + b^2 = (a-b)^2$

6. Substitute back: $x = \frac{a+b \pm \sqrt{(a-b)^2}}{2b}$

7. Simplify the square root: $x = \frac{a+b \pm |a-b|}{2b}$

Final Solution:

The two roots are:

1. When $+|a-b|$ is chosen: $x_1 = \frac{a+b + (a-b)}{2b} = \frac{2a}{2b} = \frac{a}{b}$

2. When $-|a-b|$ is chosen: $x_2 = \frac{a+b - (a-b)}{2b} = \frac{2b}{2b} = 1$

Thus, the solutions to the equation are: $x = \frac{a}{b}, \, x = 1$

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Relative Questions:

1. How is the discriminant used to determine the nature of the roots?
2. What happens to the solutions if $a = b$?
3. Can this solution be applied to specific values of $a$ and $b$?
4. How does the quadratic formula change for non-standard forms?
5. Why does the discriminant simplify to $(a-b)^2$ in this case?

Tip:

Always check the discriminant to understand whether the roots are real, equal, or complex before proceeding with further simplifications.