Virtual Math Tutor: Math problem #48

## Sunday, December 19, 2010

### Math problem #48

Math problem: It is known that $x_1$ and    $x_2$ are the roots of the quadratic equation

$ax^2 + bx + c = 0.$
Derive a quadratic equation whose roots are   $\frac{1}{x_1}$ and $\frac{1}{x_2}$.

Discussion and hints: Let us consider a general quadratic equation
$ax^2 + bx + c = 0, \quad a\not=0. \quad\quad\quad (1)$

Without loss of generality we may assume that $a >0.$ To derive the formulas for the roots of the general quadratic equation (1) we complete the square on the left hand side to get

$ax^2 + bx + c = a\left(x^2 + \frac{b}{a}x + \frac{c}{a}\right) = a\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} + \frac{c}{a} - \frac{b}{4a^2}\right) =$

$a\left[\left(x + \frac{b}{2a}\right)^2 + \frac{4ac - b^2}{4a^2}\right] = a\left(x + \frac{b}{2a}\right)^2 + \frac{4ac - b^2}{4a} =0.$

Introduce the discriminant $D= b^2 - 4ac$ of (1). It is clear from the formula above that if $D > 0$ the quadratic equation (1) has two real roots, which are given by the quadratic formula:

$x_{1,2} = \frac{-b \pm \sqrt{D}}{2a}.$

If $D=0$ then (1) admits one (double) real root

$x_1 = x_2 = - \frac{b}{2a}.$

If $D<0$ then (1) has two complex-conjugate roots

$x_{1,2} = -\frac{b}{2a} \pm i\frac{\sqrt{-D}}{2a}.$

From the formulas that we have derived above, it is easy to compute (Viète's formulas)

$x_1 + x_2 = - \frac{b}{a}, \quad x_1x_2 = \frac{c}{a}. \quad\quad (2)$

Figure 1

Let us also say a few words about maximum and minimum of the quadratic polynomial function
$y = ax^2 + bx + c.$

If $a > 0,$ then the polynomial $ax^2 + bx + c$ has no maximum value, but has the minimum of $\frac{4ac-b^2}{4a}$ at $x = -\frac{b}{2a}.$ If $a < 0,$ then the polynomial $ax^2 + bx + c$ ha no minimum, but it has the maximum value of $\frac{4ac-b^2}{4a}$ at $x = -\frac{b}{2a}.$

The graph of the function $y = ax^2 + bx + c$ is a parabola. Its vertex is at the point

$\left(-\frac{b}{2a},\frac{4ac-b^2}{4a}\right).$

If $a > 0$ the parabola is concave up, if $a < 0$ the parabola is concave down. The vertical axis of such a parabola is the line

$x = -\frac{b}{2a}.$

See Figure 1 for more details.

To complete our discussion let us consider the inequality defined by a quadratic polynomial:

$ax^2 + bx + c > 0. \quad\quad (3)$

If $a >0$ and $4ac - b^2 >0$ ($D < 0$) the inequality (3) holds true for any value of $x$ (see Figure 1). If $a < 0$ and $4ac - b^2 >0$ ($D < 0$) the inequality (3) is impossible for any value of $x$) (see Figure 1). If $a> 0$ and $4ac - b^2 < 0$ ($D>0$) the inequality (3) holds true for all values of $x$ satisfying $x< x_1$ or $x > x_2$, where $x_1, x_2\, x_1 < x_2$ are the roots of the quadratic equation (1) (see Figure 1). If $a < 0$ and $4ac - b^2 < 0$ ($D > 0$) the inequality (3) holds true for all values of $x$ satisfying $x_1 < x < x_2$, where $x_1, x_2\, x_1 < x_2$ are the roots of the quadratic equation (1) (see Figure 1). If If $a > 0$ and $4ac - b^2 = 0\, (D = 0)$ then the inequality (3) holds true for all values of If $a < 0$ and $x \in {\mathbb R}$ (except $x = -\frac{b}{2a}$, see Figure 1). If If $a < 0$ and $4ac - b^2 = 0\, (D=0)$ then the inequality (3) is impossible for any value of $a < 0$ and $x$ (see Figure 1).

Solution