The Quadratic Formula

The quadratic formula is:

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

Where $a$ , $b$ , and $c$ are the coefficients of the quadratic equation $ax^2 + bx + c = 0$ . The symbol “±” indicates that there are two possible solutions for x, one involving the addition of the square root term and the other involving its subtraction.

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Deriving the Quadratic Formula

The quadratic formula is derived by completing the square, a technique that involves rewriting a quadratic equation in the form of a perfect square trinomial. Let’s start with a general quadratic equation in the form:

$ax^2 + bx + c = 0$

Step 1: Divide both sides of the equation by $a$ :

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

Step 2: Complete the square by adding and subtracting the square of half the coefficient of the linear term, which is $\frac{b}{a}$ , not $b$ . Therefore, the correct value to use here is $\frac{(\frac{b}{a})}{2}$ . This helps create a perfect square trinomial.

To find half the coefficient of the linear term, divide the coefficient $\frac{b}{a}$ by 2, which is the same as multiplying it by $\frac{1}{2}$ :

$\frac{1}{2} \times \frac{b}{a} = \frac{b}{2a}$

Now, we’ll add and subtract the square of this value, $\left(\frac{b}{2a}\right)^2$ :

$x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 + \frac{c}{a} = 0$

Step 3: Combine the terms to form a perfect square trinomial on the left side:

$\left(x + \frac{b}{2a}\right)^2 - \frac{b^2}{4a^2} + \frac{c}{a} = 0$

Step 4: Isolate the square term by adding $\frac{b^2}{4a^2} - \frac{c}{a}$ to both sides:

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

Step 5: Combine the terms on the right side:

To combine the terms on the right side of the equation, we need a common denominator. The common denominator for the fractions is $4a^2$ , so we multiply $\frac{c}{a}</mark>$ by $4a^2$ :

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

Now, we can combine the terms:

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

Step 6: Take the square root of both sides and solve for x:

To solve for x, we need to take the square root of both sides of the equation:

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

This simplifies to:

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

Now, we can isolate x by subtracting $\frac{b}{2a}$ from both sides:

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

We can then factor out $\frac{1}{2a}$ to form the quadratic formula:

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

Solving Quadratic Equations using the Quadratic Formula

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

Step 1: Identify the coefficients $a$ , $b$ , and $c$ in the equation.

Step 2: Substitute the coefficients into the quadratic formula.

Step 3: Simplify the expression under the square root (if possible).

Step 4: Calculate the two possible values of x.

Let’s look at an example:

Solve the quadratic equation $x^2 - 5x + 6 = 0$ using the quadratic formula.

1. Identify the coefficients: $a = 1$ , $b = -5$ , and $c = 6$ .

2. Substitute the coefficients into the quadratic formula:

$x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(6)}}{2(1)}$

3. Simplify the expression under the square root:

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

4. Calculate the two possible values of x:

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

$x_1 = \frac{5 + 1}{2} = 3$

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

So, the solutions to the quadratic equation $x^2 - 5x + 6 = 0$ are $x_1 = 3$ and $x_2 = 2$ .

Let’s look at some more examples:

Examples

Question 1:

Solve the quadratic equation: $5x^2 + 3x - 2 = 0$

Solution

In this case, $a = 5$ , $b = 3$ , and $c = -2$ . Plugging these values into the quadratic formula, we get:

$x = \frac{-3 \pm \sqrt{(3)^2 - 4(5)(-2)}}{2(5)}$

$x = \frac{-3 \pm \sqrt{9 + 40}}{10}$

$x = \frac{-3 \pm \sqrt{49}}{10}$

So the two solutions for x are:

$x = \frac{-3 + 7}{10} = \frac{4}{10} = \frac{2}{5}$ and $x = \frac{-3 - 7}{10} = -1$

Question 2:

Solve the quadratic equation: $3x^2 - 2x - 1 = 0$

Solution

In this case, $a = 3$ , $b = -2$ , and $c = -1$ . Plugging these values into the quadratic formula, we get:

$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(-1)}}{2(3)}$

$x = \frac{2 \pm \sqrt{4 + 12}}{6}$

$x = \frac{2 \pm \sqrt{16}}{6}$

So the two solutions for x are:

$x = \frac{2 + 4}{6} = 1$ and $x = \frac{2 - 4}{6} = -\frac{1}{3}$

Question 3:

Solve the quadratic equation: $2x^2 - 4x - 6 = 0$

Solution

In this case, $a = 2$ , $b = -4$ , and $c = -6$ . Plugging these values into the quadratic formula, we get:

$x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(-6)}}{2(2)}$

$x = \frac{4 \pm \sqrt{16 + 48}}{4}$

$x = \frac{4 \pm \sqrt{64}}{4}$

So the two solutions for x are:

$x = \frac{4 + 8}{4} = 3$ and $x = \frac{4 - 8}{4} = -1$

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Deriving the Quadratic Formula

Solving Quadratic Equations using the Quadratic Formula

Examples

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