Example 02 - Quadratic equation problem

a = 3, b = -13, and c = -10
 

Using the quadratic formula
$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$x = \dfrac{-(-13) \pm \sqrt{(-13)^2 - 4(3)(-10)}}{2(3)}$

$x = \dfrac{13 \pm 17}{6}$

$x = 5 \, \text{ and } \, -\frac{2}{3}$   →   roots of the given equation
 

Thus, the roots of the required equation are
$x = \frac{1}{5} \, \text{ and } \, -\frac{3}{2}$
 

The required equation is
$(x - \frac{1}{5})(x + \frac{3}{2}) = 0$

$x^2 + \frac{13}{10}x - \frac{3}{10} = 0$

$10x^2 + 13x - 3 = 0$           answer

$x^2 + \dfrac{B}{A}x + \dfrac{C}{A} = 0$

$x^2 - \left( -\dfrac{B}{A} \right) x + \dfrac{C}{A} = 0$
 

From Sum and Product of Roots
$x^2 - (x_1 + x_2)x + x_1 x_2 = 0$
 

Let r₁ and r₂ = roots of the required equation
$r_1 = \dfrac{1}{x_1}$   and   $r_2 = \dfrac{1}{x_2}$
 

$r_1 + r_2 = \dfrac{1}{x_1} + \dfrac{1}{x_2}$

$r_1 + r_2 = \dfrac{x_2 + x_1}{x_1 x_2}$

$r_1 + r_2 = \dfrac{-B/A}{C/A}$

$r_1 + r_2 = -\dfrac{B}{C}$
 

$r_1 r_2 = \dfrac{1}{x_1} \cdot \dfrac{1}{x_2}$

$r_1 r_2 = \dfrac{1}{x_1 x_2}$

$r_1 r_2 = \dfrac{1}{C/A}$

$r_1 r_2 = \dfrac{A}{C}$
 

The required equation is:
$x^2 - (r_1 + r_2)x + r_1 r_2 = 0$

$x^2 - \left( -\dfrac{B}{C} \right)x + \dfrac{A}{C} = 0$

$x^2 + \dfrac{B}{C}x + \dfrac{A}{C} = 0$

$Cx^2 + Bx + A = 0$   ^†
 

From the given equation: A = 3, B = -13, and C = -10
$-10x^2 - 13x + 3 = 0$

$10x^2 + 13x - 3 = 0$           answer
 

^† The quadratic equation whose roots are reciprocals to the roots of Ax² + Bx + C = 0 is Cx² + Bx + A = 0.