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.

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

• 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. 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

## 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. ### How To Find The Factors Of 20: A Simple Way

Sep 17, 23 09:46 AM

There are many ways to find the factors of 20. A simple way is to...