# The Discriminant and Complex Roots

The expression ${b^2} – 4ac$, which appears under the radical sign in the quadratic formula
$x = \frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}$

is called the discriminant of the quadratic equation $a{x^2} + bx + c = 0$.

If  $a$, $b$ and $c$ are real numbers, you can use the algebraic sign of the discriminant to determine the number and the nature of the roots of the quadratic equation.

1. If ${b^2} – 4ac$ is positive, the equation has two real and unequal roots.
2. If ${b^2} – 4ac = 0$, the equation has only one root, i.e., double roots.
3. If ${b^2} – 4ac$ is negative, the equation has no real root. Its roots are two complex numbers that are complex conjugates of each other.

Example:

Use the discriminant to determine the nature of the roots of each quadratic equation without actually solving it.

(a) $5{x^2} – x – 3 = 0$

(b) $9{x^2} + 42x + 49 = 0$

(c) ${x^2} – x + 1 = 0$

Solution:

(a) Here $a = 5$, $b = – 1$, $c = – 3$ and ${b^2} – 4ac = {\left( { – 1} \right)^2} – 4\left( 5 \right)\left( { – 3} \right) = 61$ is positive, hence there are two unequal real roots.

(b) Here $a = 9$, $b = 42$, $c = 49$ and ${b^2} – 4ac = {\left( {42} \right)^2} – 4\left( 9 \right)\left( {49} \right) = 0$, hence the equation has just one root: a double root and this root is a real number.

(c) Here $a = 1$, $b = – 1$, $c = 1$ and ${b^2} – 4ac = {\left( { – 1} \right)^2} – 4\left( 1 \right)\left( 1 \right) = – 3$ is negative, hence the equation has no real root.

Example:

Use the quadratic formula to find the roots of the quadratic equation ${x^2} – x + 1 = 0$.

Solution:

Using the quadratic formula, with $a = 1$, $b = – 1$ and $c = 1$, we have
$x = \frac{{ – b \pm \sqrt {{b^2} – 4ac} }}{{2a}}$
$x = \frac{{ – \left( { – 1} \right) \pm \sqrt {{{\left( { – 1} \right)}^2} – 4\left( 1 \right)\left( 1 \right)} }}{{2\left( 1 \right)}}$
$= \frac{{1 \pm \sqrt { – 3} }}{2}$
$= \frac{{1 \pm \sqrt 3 i}}{2} = \frac{1}{2} \pm \frac{{\sqrt 3 }}{2}i$

Thus, the roots are the complex conjugates $\frac{1}{2} + \frac{{\sqrt 3 }}{2}i$ and $\frac{1}{2} – \frac{{\sqrt 3 }}{2}i$