Sub Topics

In Algebra help, An equation is an equality which is satisfied only by some particular values of the variables occurring in it. Solving equations is a very important topic covered under algebra.

An equation of the form ax² + bx + c = 0 where a, b, c are certain numbers, and a $\neq$ 0 is called a quadratic equations. The number a, b, c are called the coefficients of the quadratic equation and the number b² - 4ac is called its discriminant. Discriminant of a quadratic equation is usually denoted by D.

A quadratic equation in many cases can be factored in the form

(x – x1)(x – x2) = 0, where x1 and x2 are the roots, and the roots can be found by equating each factor to 0. But also in many cases factoring found to be not possible. In such situations, a formula has been derived to find the roots in general. It states that, if ax2 + bx + c = 0 and if x1 and x2 are the roots, then,

$x_1$ = $\frac{-b + \sqrt{b^2-4ac}}{2a}$

$x_2$ = $\frac{-b - \sqrt{b^2-4ac}}{2a}$

The term b2 – 4ac is called as the discriminant, D, of the quadratic equation and it decides the nature of the root. Rewriting the roots in discriminant form,

$x_1$ = $\frac{-b + \sqrt{D}}{2a}$

$x_2$ = $\frac{-b - \sqrt{D}}{2a}$

Now, we find the followings.

If D < 0, the radicand is negative and $\sqrt{D}$ is imaginary and has two values. Therefore, if D < 0, the roots are imaginary and different.

If D = 0, the radicand is 0 and $\sqrt{D}$ becomes 0. Therefore, if D = 0, the roots are real and identical.

If D > 0, the radicand is positive and $\sqrt{D}$ is real and has two values. Therefore, if D > 0, the roots are real and different.

We have seen that the roots of a quadratic equation are,

$x_1$ = $\frac{-b + \sqrt{b^2-4ac}}{2a}$

$x_2$ = $\frac{-b - \sqrt{b^2-4ac}}{2a}$

$x_1 + x_2$ = $\frac{-b + \sqrt{b^2-4ac}}{2a}$ + $\frac{-b - \sqrt{b^2-4ac}}{2a}$ = $\frac{-b}{a}$
Therefore, the sum of the two roots is $\frac{-b}{a}$ and this is also the x-coordinate of the vertex of the function. Also, therefore, x = (-b/a) represents the equation of symmetry.
$x_1 . x_2$ = $\frac{-b + \sqrt{b^2-4ac}}{2a}$ $\frac{-b - \sqrt{b^2-4ac}}{2a}$ = $\frac{c}{a}$
Therefore, the product of the two roots is $\frac{c}{a}$.