Quadratic Formula

What is the Quadratic Formula?

A quadratic equation is an equation in the form $ax^2 + bx + c = 0$ where $a \neq 0$, where:

• $a$ 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:

This formula will give us two values of $x$ which are the roots of the quadratic equation:

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

Deriving the Quadratic Formula

The quadratic formula can be derived from the method of completing the square. Let's see how:

Starting with the standard form of a quadratic equation:

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 $x$:

Thus, 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

Examples

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:

Therefore, 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:

Therefore, 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 discriminant provides information about the nature of the 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:

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 we know the roots of a quadratic equation, we can create a new quadratic equation. Suppose $p$ and $q$ are the roots of a quadratic equation, then the equation is:

Or in standard form:

Application Examples

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

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

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

   Step 3: Create the new quadratic equation

   $$x^2 - \frac{4}{21}x - \frac{1}{84} = 0$$

Practice Problems

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).

   Therefore, the root of the equation is $x_1 = x_2 = -\frac{1}{6}$.

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 $x_2$:

   $$x_2 = -4 - 2 = -6$$

   Therefore, 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 $x_2$:

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

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