Types of Quadratic Equation Roots

Quadratic Equation Roots

In a quadratic equation $ax^2 + bx + c = 0$ with $a \neq 0$ , a root is any value of $x$ that satisfies the equation: plug it in and both sides match. For example, $x = 2$ in $x^2 - 5x + 6 = 0$ gives $4 - 10 + 6 = 0$ , so 2 is a root. Over the real numbers alone, an equation with a negative discriminant has no root. But over the complex numbers, every quadratic equation has two roots. When both roots are the same, as in $x^2 - 6x + 9 = 0$ below, we count that repeated root twice. The quadratic formula computes both roots directly from $a$ , $b$ , and $c$ :

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

The expression $b^2 - 4ac$ is the discriminant, denoted by $D$ . When $D > 0$ the roots are two distinct reals. When $D = 0$ the root is repeated. When $D < 0$ the roots are nonreal complex. So the sign of $D$ already tells you the root type before you finish evaluating the formula.

Different Real Roots

If $D > 0$ (or $b^2 > 4ac$ ), then the quadratic equation has two different real roots.

\begin{aligned} &x_1 = \frac{-b + \sqrt{D}}{2a} \\ &\text{and} \\ &x_2 = \frac{-b - \sqrt{D}}{2a} \end{aligned}

Example: $x^2 - 5x + 6 = 0$

$a = 1, b = -5, c = 6$
$D = (-5)^2 - 4 \cdot 1 \cdot 6 = 25 - 24 = 1 > 0$
The roots are: $x_1 = 3$ and $x_2 = 2$

Equal or Repeated Roots

If $D = 0$ (or $b^2 = 4ac$ ), then the quadratic equation has one real repeated root (two roots with the same value).

x_1 = x_2 = \frac{-b}{2a}

Example: $x^2 - 6x + 9 = 0$

$a = 1, b = -6, c = 9$
$D = (-6)^2 - 4 \cdot 1 \cdot 9 = 36 - 36 = 0$
The roots are: $x_1 = x_2 = 3$

Two Nonreal Complex Roots

If $D < 0$ (or $b^2 < 4ac$ ), then the quadratic equation has two distinct non-real complex roots. For real coefficients, they form a conjugate pair.

\begin{aligned} &x_1 = \frac{-b + i\sqrt{|D|}}{2a} \\ &\text{and} \\ &x_2 = \frac{-b - i\sqrt{|D|}}{2a} \end{aligned}

Here $i^2=-1$ defines the imaginary unit.

Example: $x^2 + 2x + 5 = 0$

$a = 1, b = 2, c = 5$
$D = 2^2 - 4 \cdot 1 \cdot 5 = 4 - 20 = -16 < 0$
The roots are: $x_1 = -1 + 2i$ and $x_2 = -1 - 2i$

Relationship Between Roots and Coefficients

If $x_1$ and $x_2$ are the roots of the quadratic equation $ax^2 + bx + c = 0$ , then:

x_1 + x_2 = -\frac{b}{a}

These relationships can verify calculated roots or determine the coefficient ratios $b/a$ and $c/a$ . A specific equation is fixed once the leading coefficient $a$ is also known.

Exercise

The problem asks you to classify first and solve second, so compute the discriminant before taking any square root. Checking the sum and the product afterwards shows whether the two roots belong together.

Classify and solve the equation $2x^2-4x+5=0$ . Then verify the roots using their sum and product.

Worked Solution

The coefficients are $a=2$ , $b=-4$ , and $c=5$ . Its discriminant is

\begin{aligned} D &=(-4)^2-4\cdot2\cdot5 \\ &=16-40 \\ &=-24 \end{aligned}

Because $D<0$ , the equation has two non-real complex roots. Apply the quadratic formula and use $\sqrt{-24}=2i\sqrt6$ :

x=\frac{-(-4)\pm\sqrt{-24}}{2\cdot2}

The calculation gives:

\begin{aligned} &x_1=1+\frac{i\sqrt6}{2} \\ &\text{and} \\ &x_2=1-\frac{i\sqrt6}{2} \end{aligned}

Their sum and product agree with Vieta's formulas:

x_1+x_2=2=-\frac{b}{a}

Both checks use the original coefficients, so they also catch a lost sign or an incorrect denominator in the quadratic formula.