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

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