Quadratic Formula

Calculating Roots from Quadratic Coefficients

A quadratic equation has the form $ax^2 + bx + c = 0$, where $a \neq 0$, with the following coefficients:

• $a$ 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:

Over the real numbers alone, an equation with a negative discriminant has no root. But over the complex numbers, the two signs give two roots. When the discriminant is zero, both expressions below give the same repeated root, and we count that repeated root twice:

• $x_1 = \frac{-b + \sqrt{b^2 - 4ac}}{2a}$ (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:

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

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

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

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

Step 5: Simplify the right side:

Step 6: Take the square root of both sides:

Step 7: Solve for $x$:

We obtain the quadratic formula:

Using the Quadratic Formula

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

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

For $x_1$, take the positive sign:

For $x_2$, take the negative sign:

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:

For $x_1$, take the positive sign:

For $x_2$, take the negative sign:

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

The sum and the product of the two roots can be read off the coefficients without solving the equation first. That gives you a quick check on a solution or a way to find a missing value.

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

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

Proving the Relationships

From the quadratic formula, we know that:

Adding the roots:

Multiplying the roots:

Creating New Quadratic Equations from Known Roots

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

Or in standard form:

Changing the Roots of a Quadratic Equation

1. 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 $p \cdot q$

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

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

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

   $$2p + 2q = 2(p + q) = 2(-4) = -8$$

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

   Step 3: Create the new quadratic equation

   $$x^2 - (-8)x + 12 = 0$$

   $$x^2 + 8x + 12 = 0$$

2. The quadratic equation $x^2 + 4x - 21 = 0$ has roots $p$ and $q$.

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

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

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

   $$p \cdot q = \frac{c}{a} = \frac{-21}{1} = -21$$

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

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

   $$\begin{aligned} \frac{1}{2p} \cdot \frac{1}{2q} &= \frac{1}{4pq} \\ &= \frac{1}{4 \cdot (-21)} \\ &= \frac{1}{-84} \\ &= -\frac{1}{84} \end{aligned}$$

   Step 3: Create the new quadratic equation

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

Practice Problems

Each equation is already in standard form, so read the coefficients before you substitute them into the formula. Check the value under the square root first to see how many real solutions to expect.

Solve the following quadratic equations using the quadratic formula:

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

Answer Key

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

   Identify: $a = 1$, $b = 5$, $c = 6$

   $$x_{1,2} = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1}$$

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

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

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

   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 $x_2 = -3$.

2. Solution to the quadratic equation $2x^2 + 6x + 3 = 0$

   Identify: $a = 2$, $b = 6$, $c = 3$

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

   $$= \frac{-6 \pm \sqrt{36 - 24}}{4}$$

   $$= \frac{-6 \pm \sqrt{12}}{4}$$

   $$= \frac{-6 \pm 2\sqrt{3}}{4}$$

   $$= \frac{-3 \pm \sqrt{3}}{2}$$

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

3. Solution to the quadratic equation $6x^2 + 2x + \frac{1}{6} = 0$

   Identify: $a = 6$, $b = 2$, $c = \frac{1}{6}$

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

   $$= \frac{-2 \pm \sqrt{4 - 4}}{12}$$

   $$= \frac{-2 \pm 0}{12}$$

   $$= \frac{-2}{12} = -\frac{1}{6}$$

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

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

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

   Identify: $a = \frac{1}{2}$, $b = 4$, $c = 6$

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

   $$= \frac{-4 \pm \sqrt{16 - 12}}{1}$$

   $$= \frac{-4 \pm \sqrt{4}}{1}$$

   $$= -4 \pm 2$$

   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 $x_2 = -6$.

5. Solution to the quadratic equation $\frac{2}{3}x^2 + 2x - 12 = 0$

   Identify: $a = \frac{2}{3}$, $b = 2$, $c = -12$

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

   $$= \frac{-2 \pm \sqrt{4 + 32}}{4/3}$$

   $$= \frac{-2 \pm \sqrt{36}}{4/3}$$

   $$= \frac{-2 \pm 6}{4/3}$$

   $$= \frac{3(-2 \pm 6)}{4}$$

   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 $x_2 = -6$.