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.

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

In this case, a = 5a = 5, b = 3b = 3, and c = -2c = -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{(3)^2 - 4(5)(-2)}}{2(5)}

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

x = \frac{-3 \pm \sqrt{49}}{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}x = \frac{-3 + 7}{10} = \frac{4}{10} = \frac{2}{5} and x = \frac{-3 - 7}{10} = -1x = \frac{-3 - 7}{10} = -1

Question 2:

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

In this case, a = 3a = 3, b = -2b = -2, and c = -1c = -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{(-2)^2 - 4(3)(-1)}}{2(3)}

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

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

So the two solutions for x are:

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

Question 3:

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

In this case, a = 2a = 2, b = -4b = -4, and c = -6c = -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{(-4)^2 - 4(2)(-6)}}{2(2)}

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

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

So the two solutions for x are:

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

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