# Quadratic Equations: Learn and Crack CAT Quantitative Aptitude

If you analyze the past CAT exam papers you will realize that invariably two to three problems involve the application of Quadratic Equations and Inequations. This article lays the foundation of the concept.  Out of complete section on Quantitative Aptitude, two or three questions are covered from Quadratic Equations every year.

A quadratic equation is an equation of degree two. The general form of such an equation is therefore
ax2 + bx + c = 0,         (1)

…where a, b, c are any constants, typically real or complex numbers. We need to assume where that a≠0; otherwise, the degree of the polynomial ax2 + bx + c = bx + c would not remain two. We shall, therefore, assume that a≠0 throughout this section.

There are two aspects to a quadratic.

1. We could view this geometrically, as a function f(x) = ax2 + bx + c or simply as
y = ax2 + bx + c,

2. Or we could think of this in algebraic terms, as an equation to solve.

If we try to plot the graph of the function f(x) = ax2 + bx + c, where a≠0, we find that we always trace a parabola.This is easily seen by “completing squares”, a technique you may find useful in several situations. Let us apply this procedure to 4af(x):

4af(x) = 4a2x2 + 4abx + 4ac = (2ax + b)2 – b2 + 4ac = (2ax + b)2 - Δ,     (2)

…where Δ = b2 - 4ac is called the discriminant of the polynomial and plays a crucial role in determining the nature of the solutions of an equation such as (1).

The algebraic aspect is perhaps more useful in the context of this article. We will undertake the geometrical aspect in another article.

If we are given a quadratic equation, such as (1), we would like to be able to:

(a) determine if the equation has a solution; and (b) to find the solution, in case the answer is yes to part (a).

Finding a solution (or, finding a “root” of equation (1)) is finding a real (or complex) number x0 which meets the condition:

ax20+bx0+c = 0

Any solution x0 of equation (1) must also satisfy 4af(x0) = 0 (and, in fact, any solution of 4af(x) = 0 must also be a solution of f(x) = 0 since a  0). If you now look at equation (2), you will find that it can be re-written as:

(2ax0 + b)2 = Δ        (3)

a. If the numbers a, b, c are all real and you are looking for real x0, equation (3) can hold only if Δ = b2 - 4ac ≥ 0, since only non-negative (real) numbers have a (real) square root.

b. If you allow for non-real solutions as well, this restriction on the sign of Δ is not only removed, but one may also allow a, b, c to be non-real. With this understanding, first taking square roots, then subtracting b from each side and finally dividing both sides by the non-zero number 2a, we get:

(4)

Quadratice Equation is a concept introduced in smaller classes and once introduced, it gets applied in questions at every level. Being conceptually clear is very important.

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