Quadratic Formula

a \neq 0

$a$ is the coefficient of $x^2$
$b$ is the coefficient of $x$
$c$ is the constant term

To solve a quadratic equation $ax^2 + bx + c = 0$ , we can use the formula:

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

Over the complex numbers, the two signs give the two roots counted with multiplicity. When the discriminant is zero, both expressions give the same repeated root:

$x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$ (using the plus sign)
$x_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a}$ (using the minus sign)

The part $b^2 - 4ac$ is called the discriminant and determines the nature of the roots:

If $b^2 - 4ac > 0$ : Two distinct real roots
If $b^2 - 4ac = 0$ : One real root (a repeated root)
If $b^2 - 4ac < 0$ : No real roots, but two non-real complex conjugate roots

Deriving the Quadratic Formula

To derive the quadratic formula, complete the square in the general form below:

ax^2 + bx + c = 0

Step $1$ : Divide all terms by $a$ (the coefficient of $x^2$ ):

x^2 + \frac{b}{a}x + \frac{c}{a} = 0

Step $2$ : Move the constant term to the right side:

x^2 + \frac{b}{a}x = -\frac{c}{a}

Step $3$ : Add the square of half the coefficient of $x$ to both sides:

x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} = -\frac{c}{a} + \frac{b^2}{4a^2}

Step $4$ : The left side now forms a perfect square:

\left(x + \frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} - \frac{c}{a}

Step $5$ : Simplify the right side:

\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}

Step $6$ : Take the square root of both sides:

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

Step $7$ : Solve for $x$ :

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

We obtain the quadratic formula:

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

Using the Quadratic Formula

To solve a quadratic equation using the formula, follow these steps:

Make sure the quadratic equation is in standard form $ax^2 + bx + c = 0$
Identify the values of $a$ , $b$ , and $c$
Substitute these values into the formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Calculate the values of $x$ to find the roots of the equation

Solving with the Quadratic Formula

Example $1$ : Solve the equation $x^2 + 5x + 6 = 0$

Identify the values: $a = 1$ , $b = 5$ , and $c = 6$

Substitute into the formula:

x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1} = \frac{-5 \pm \sqrt{25 - 24}}{2} = \frac{-5 \pm \sqrt{1}}{2} = \frac{-5 \pm 1}{2}

For $x_1$ , take the positive sign:

x_1 = \frac{-5 + 1}{2} = \frac{-4}{2} = -2

For $x_2$ , take the negative sign:

x_2 = \frac{-5 - 1}{2} = \frac{-6}{2} = -3

The roots of the equation are $x_1 = -2$ and $x_2 = -3$

Example $2$ : Solve the equation $2x^2 - 7x + 3 = 0$

Identify the values: $a = 2$ , $b = -7$ , and $c = 3$

Substitute into the formula:

x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \cdot 2 \cdot 3}}{2 \cdot 2} = \frac{7 \pm \sqrt{49 - 24}}{4} = \frac{7 \pm \sqrt{25}}{4} = \frac{7 \pm 5}{4}

For $x_1$ , take the positive sign:

x_1 = \frac{7 + 5}{4} = \frac{12}{4} = 3

For $x_2$ , take the negative sign:

x_2 = \frac{7 - 5}{4} = \frac{2}{4} = \frac{1}{2}

The roots of the equation are $x_1 = 3$ and $x_2 = \frac{1}{2}$

The Discriminant of a Quadratic Equation

The expression $b^2 - 4ac$ in the quadratic formula is called the discriminant, often denoted by $D$ or $\Delta$ .

The sign of the discriminant determines the type of roots of a quadratic equation:

If $D > 0$ : The equation has two distinct real roots
If $D = 0$ : The equation has one real root (a repeated root)
If $D < 0$ : The equation has no real roots (the roots are complex numbers)

Relationship Between Roots and Coefficients

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

Sum of the roots: $x_1 + x_2 = -\frac{b}{a}$
Product of the roots: $x_1 \cdot x_2 = \frac{c}{a}$

Proving the Relationships

From the quadratic formula, we know that:

x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \quad \text{and} \quad x_2 = \frac{-b - \sqrt{b^2 - 4ac}}{2a}

Adding the roots:

x_1 + x_2 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} + \frac{-b - \sqrt{b^2 - 4ac}}{2a} = \frac{-2b}{2a} = -\frac{b}{a}

Multiplying the roots:

x_1 \cdot x_2 = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \cdot \frac{-b - \sqrt{b^2 - 4ac}}{2a} = \frac{b^2 - (\sqrt{b^2 - 4ac})^2}{4a^2} = \frac{b^2 - (b^2 - 4ac)}{4a^2} = \frac{4ac}{4a^2} = \frac{c}{a}

Creating New Quadratic Equations from Known Roots

If $p$ and $q$ are known roots, one monic quadratic equation with those roots is:

(x - p)(x - q) = 0

Or in standard form:

x^2 - (p+q)x + pq = 0

Changing the Roots of a Quadratic Equation

The quadratic equation $x^2 + 4x + 3 = 0$ has roots $p$ and $q$ .

Find the quadratic equation with roots $2p$ and $2q$ .

Step $1$ : Find the values of $p + q$ and $p \cdot q$
$p + q = -\frac{b}{a} = -\frac{4}{1} = -4$
Step $2$ : Calculate the sum and product of the new roots
$2p + 2q = 2(p + q) = 2(-4) = -8$
Step $3$ : Create the new quadratic equation
$x^2 - (-8)x + 12 = 0$
The quadratic equation $x^2 + 4x - 21 = 0$ has roots $p$ and $q$ .

Practice Problems

Solve the following quadratic equations using the quadratic formula:

$x^2 + 5x + 6 = 0$
$2x^2 + 6x + 3 = 0$
$6x^2 + 2x + \frac{1}{6} = 0$
$\frac{1}{2}x^2 + 4x + 6 = 0$
$\frac{2}{3}x^2 + 2x - 12 = 0$

Answer Key

Solution to the quadratic equation $x^2 + 5x + 6 = 0$

Identify: $a = 1$ , $b = 5$ , $c = 6$
$x_{1,2} = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1}$
For $x_1$ :

$x_1 = \frac{-5 + 1}{2} = \frac{-4}{2} = -2$

For $x_2$ :

$x_2 = \frac{-5 - 1}{2} = \frac{-6}{2} = -3$

Therefore, the roots of the equation are $x_1 = -2$ and $x_2 = -3$ .
Solution to the quadratic equation $2x^2 + 6x + 3 = 0$

Identify: $a = 2$ , $b = 6$ ,
Solution to the quadratic equation $6x^2 + 2x + \frac{1}{6} = 0$

Identify: , ,
Solution to the quadratic equation $\frac{1}{2}x^2 + 4x + 6 = 0$

Identify: , ,
Solution to the quadratic equation $\frac{2}{3}x^2 + 2x - 12 = 0$

Identify: , ,

p \cdot q = \frac{c}{a} = \frac{3}{1} = 3

2p \cdot 2q = 4(p \cdot q) = 4 \cdot 3 = 12

Find the quadratic equation with roots $\frac{1}{2p}$ and $\frac{1}{2q}$ .

Step $1$ : Find the values of $p + q$ and $p \cdot q$

p + q = -\frac{b}{a} = -\frac{4}{1} = -4

Step $2$ : Calculate the sum and product of the new roots

\frac{1}{2p} + \frac{1}{2q} = \frac{1}{2} \left(\frac{q + p}{pq}\right) = \frac{1}{2} \cdot \frac{-4}{-21} = \frac{1}{2} \cdot \frac{4}{21} = \frac{4}{42} = \frac{2}{21}

Step $3$ : Create the new quadratic equation

x^2 - \frac{2}{21}x - \frac{1}{84} = 0

= \frac{-5 \pm \sqrt{25 - 24}}{2}

= \frac{-5 \pm \sqrt{1}}{2}

= \frac{-5 \pm 1}{2}

c = 3

x_{1,2} = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 2 \cdot 3}}{2 \cdot 2}

For $x_1$ :

x_1 = \frac{-3 + \sqrt{3}}{2}

For $x_2$ :

x_2 = \frac{-3 - \sqrt{3}}{2}

Therefore, the roots of the equation are $x_1 = \frac{-3 + \sqrt{3}}{2}$ and $x_2 = \frac{-3 - \sqrt{3}}{2}$ .

a = 6

b = 2

c = \frac{1}{6}

x_{1,2} = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 6 \cdot \frac{1}{6}}}{2 \cdot 6}

Since the discriminant $b^2 - 4ac = 0$ , the equation has one root (a repeated root).

The equation has the repeated root $x_1 = x_2 = -\frac{1}{6}$ .

a = \frac{1}{2}

b = 4

c = 6

x_{1,2} = \frac{-4 \pm \sqrt{4^2 - 4 \cdot \frac{1}{2} \cdot 6}}{2 \cdot \frac{1}{2}}

For $x_1$ :

x_1 = -4 + 2 = -2

For $x_2$ :

x_2 = -4 - 2 = -6

The roots of the equation are $x_1 = -2$ and $x_2 = -6$ .

a = \frac{2}{3}

b = 2

c = -12

x_{1,2} = \frac{-2 \pm \sqrt{2^2 - 4 \cdot \frac{2}{3} \cdot (-12)}}{2 \cdot \frac{2}{3}}

For $x_1$ :

x_1 = \frac{3(-2 + 6)}{4} = \frac{3 \cdot 4}{4} = 3

For $x_2$ :

x_2 = \frac{3(-2 - 6)}{4} = \frac{3 \cdot (-8)}{4} = -6

The roots of the equation are $x_1 = 3$ and $x_2 = -6$ .

Quadratic Formula

Calculating Roots from Quadratic Coefficients

Quadratic Formula

Calculating Roots from Quadratic Coefficients

Deriving the Quadratic Formula

Using the Quadratic Formula

Solving with the Quadratic Formula

The Discriminant of a Quadratic Equation

Relationship Between Roots and Coefficients

Proving the Relationships

Creating New Quadratic Equations from Known Roots

Changing the Roots of a Quadratic Equation

Practice Problems

Answer Key