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 

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

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.

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.

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