**Problem**

In a quadratic equation problem, one student made a mistake in copying the coefficient of x and got roots of 3 and -2. Another student made a mistake in copying the constant term and got the roots of 3 and 2. What are the correct roots?

**Solution**

$ax^2 + bx + c = 0$

For the first student, b is wrong but a and c are correct. Thus, the product of roots is the same as that of the correct equation.

$x_1 x_2 = \dfrac{c}{a}$

$3(-2) = \dfrac{c}{a}$

$\dfrac{c}{a} = -6$

For the second student, c is wrong but a and b are correct. Thus, the sum of roots is the same as that of the correct equation.

$x_1 + x_2 = -\dfrac{b}{a}$

$3 + 2 = -\dfrac{b}{a}$

$\dfrac{b}{a} = -5$

From the correct equation

$ax^2 + bx + c = 0$

$x^2 + \dfrac{b}{a}x + \dfrac{c}{a} = 0$

$x^2 - 5x - 6 = 0$ → the correct equation

Solving for the correct roots

$x^2 - 5x - 6 = 0$

$(x - 6)(x + 1) = 0$

$x = 6 \, \text{ and } \, -1$ → the correct roots *answer*

**Another Solution**

$(x - 3)(x + 2) = 0$

$x^2 - x - 6 = 0$

a = 1 and c = -6 are correct, b = -1 is wrong

Second student: roots are 3 and 2

$(x - 3)(x - 2) = 0$

$x^2 - 5x + 6 = 0$

a = 1 and b = -5 are correct, c = 6 is wrong

Thus, a = 1, b = -5, and c = -6. The correct equation is

$ax^2 + bx + c = 0$

$x^2 - 5x - 6 = 0$

$(x - 6)(x + 1) = 0$

$x = 6 \, \text{ and } \, -1$ → the correct roots *answer*