Types of Roots of Quadratic Equations

2.2

Types of Roots of Quadratic Equations

A diagram showing the quadratic equation formula and its components.

Introduction

The type of roots of a quadratic equation \(ax^2+bx+c=0\) can be determined by finding the value of discriminant, \(D=b^2-4ac\).

Condition of \(D\)	Type of Roots
\(b^2-4ac \gt 0\)	Two different real roots
\(b^2-4ac=0\)	Two equal real roots
\(b^2-4ac \lt 0\)	No real roots

Example \(1\)

Question

Determine the type of roots for the quadratic equation \(x^2+5x-6=0\).

Solution

Based on equation \(x^2+5x-6=0\),

\(a=1\),
\(b=5\),
\(c=-6\).

Use the formula for discriminant,

\(\begin{aligned} b^2-4ac&=5^2-4(1)(-6) \\ &=49 (\gt0) .\end{aligned}\)

Thus, the equation \(x^2+5x-6=0\) has two real and different roots.

Example \(2\)

Question

The quadratic equation \(x^2+k+3=kx\), where \(k\) is a constant, has two equal real roots. Find the possible values of \(k\).

Solution

Arrange the equation in general form,

\(\begin{aligned} x^2+k+3&=kx\\ x^2-kx+k+3&=0. \end{aligned}\)

From the equation,

\(a=1\),
\(b=-k\),
\(c=k+3\).

Use the formula of discriminant,

\(\begin{aligned} b^2-4ac&=0\\ (-k)^2-4(1)(k+3)&=0\\ k^2-4k-12&=0\\ (k+2)(k-6)&=0 \end{aligned}\)

\(k=-2\) or \(k=6\).

Thus, the possible values for \(k\) are \(-2\) or \(6\).