# Quadratic formula

The quadratic formula is a math formula that can be used to solve a quadratic equation that is written in standard form ax2 + bx + c = 0

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

Notice that the general form of a quadratic equation is ax2 + bx + c = 0 and it is a second order equation in a single variable. Before using the formula shown above, it is important to check two things:

• First, make sure that the coefficient of the leading term is not equal to zero (a ≠ 0). If a is equal to zero, then the equation becomes a linear equation instead.
• Then, make sure that the quadratic equation is indeed written in standard or general form.

The plus or minus sign (±) in the formula is there to show that the quadratic equation may have two solutions (x1 and x2) generally speaking.

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

## Important definitions about the quadratic formula

• The expression inside the radical symbol or square root sign is called radicand. Therefore, the radicand in the quadratic formula is b2 - 4ac.
• The discriminant of a quadratic equation in the form ax2 + bx + c = 0 is the value of the expression b2 - 4ac.
• When you are dealing with a function, x1 and x2 are usually called zeros of the function.
• When you are dealing with a quadratic equation, x1 and x2 are usually called solutions or roots of the quadratic equation.
• When you are graphing a quadratic function, (x1, 0) and and (x2, 0) are called x-intercepts since these are points where the parabola crosses the x-axis.

## The Discriminant

The expression b2 - 4ac, called discriminant, shows the nature of the solutions.

It is common to use the symbol Δ (delta from the Greek alphabet) when talking about the discriminant.

Δ = b2 - 4ac

If Δ or the discriminant is zero, then it makes no difference whether we choose the plus or the minus sign in the formula.

x1 = x2 = -b/2a

In this case, we say that there is one repeated real solution.

If Δ or the discriminant is positive, then there will be two real solutions.

If Δ or the discriminant is negative, then we will end up taking the square root of a negative number. In this case, there will be two imaginary-number solutions called complex numbers.

Discriminant

For ax2 + bx + c = 0:

Δ = b2 - 4ac = 0  One real-number solution

Δ = b2 - 4ac > 0  Two different real-number solutions

Δ = b2 - 4ac < 0  No real solution, but two different imaginary-number solutions.

Check the lesson about discriminant of the quadratic equation to see what the graph of a quadratic equation looks like when the discriminant is either zero, positive, or negative.

## Using the quadratic formula to solve a quadratic equation

Solve x2 - 5x + 4 = 0 using the quadratic formula

a = 1, b = -5, and c = 4

$$x = \frac{-(-5) ± \sqrt{(-5)^2 - 4(1)(4)}}{2(1)}$$ $$x = \frac{5 ± \sqrt{25 - 16}}{2}$$ $$x = \frac{5 ± \sqrt{9}}{2}$$ $$x = \frac{5 ± 3}{2}$$ $$x_1 = \frac{5 + 3}{2}\ and\ x_2 = \frac{5 - 3}{2}$$ $$x_1 = \frac{8}{2}\ and\ x_2 = \frac{2}{2}$$ $$x_1 = 4\ and\ x_2 = 1$$

The roots of the equation x2 - 5x + 4 = 0 are x1 = 4 and x2 = 1

Please check the lesson about solve using the quadratic formula to see more examples.

## Applications

Example #1

Suppose a soccer player shoots a penalty kick with an initial velocity of 28 ft/s. When will the ball reach a height of 30 feet?

Solution

The function h = -16t2 + vt + s models the height h in feet of the ball at time t in seconds.

The velocity is v and s is the initial height of the ball.

Since the soccer ball must be on the ground before the soccer player shoots the ball, s is equal to 0.

v = 28 ft/s

h is the height of the ball

30 = -16t2 + 28t

Since the standard form of a quadratic equation is ax2 + bx + c = 0, you need to put 30 = -16t2 + 28t in standard form.

Subtract 30 from both sides of the equation

30 - 30 = -16t2 + 28t - 30

0 = -16t2 + 28t - 30

-16t2 + 28t - 30 = 0

Find the values of a,b, and c and then evaluate the discriminant.

a = -16, b = 28 and c = -30

Δ = b2 - 4ac  = 282 - 4(-16)(-30)

Δ = b2 - 4ac  = 784 + 64(-30)

Δ = b2 - 4ac  = 784 + -1920

Δ = b2 - 4ac  = -1136

Since the discriminant is negative, the equation 30 = -16t2 + 28t has no real solutions.

Therefore, the ball will not reach a height of 30 feet.

Example #2

Find the dimensions of a square that has the same area as a circle whose radius is 10 inches.

Solution

Let x be the length of one side of the square. Then, the area of the square is x2

The area of the circle is pir2 = 3.14(10)2 = 3.14(100) = 314

x2 = 314

x2 - 314 = 0

x2 - 0x - 314 = 0

a = 1, b = 0, and c = -314

Δ = b2 - 4ac  = 02 - 4(1)(-314)

Δ = b2 - 4ac  =  1256

√Δ = √(1256) = 35.44

x1 = (-b + 35.44) / 2(1)

x1 = (-0 + 35.44) / 2

x1 = 35.44 / 2 = 17.72

x2 = (-b  - 35.44) / 2(1)

x2 = (-0 - 35.44) / 2

x2 = -35.44 / 2 = -17.72

The length of one side of the square is 17.72 inches

## Recent Articles

1. ### Rational Numbers - Definition and Examples

Mar 15, 23 07:45 AM

To learn about rational numbers, write their decimal expansion, and recognize rational numbers that are repeating decimals and terminating decimals.

Read More

2. ### Area of a Trapezoid - Definition, Formula, and Examples

Mar 13, 23 07:52 AM

Learn how to get the area of a trapezoid using a rectangle and a triangle, the formula, and also when the height of the trapezoid is missing.

Read More

100 Tough Algebra Word Problems.

If you can solve these problems with no help, you must be a genius!

Recommended