Unlocking Solving Quadratic Equations Using the Quadratic Formula

Here are a few recommended readings before getting started with this lesson.

Square Root
Quadratic Equation
Standard Form of a Quadratic Function
Completing the Square

Try a few practice exercises as a warm-up!

4^{2} - 4 (1) (2)

by using the properties of square roots to remove the perfect-square factor.

b Consider the quadratic function

y = (x + 2)^{2} - 1 .

Of the following expressions, which represents the same function written in standard form?

c Identify the coefficient

b

of the quadratic equation

x - 5 = 2 (x - 3)^{2}

when written in standard form

a x^{2} + b x + c = 0 .

$x = \frac{- 3 2 \pm 1 6}{- 4}$
$x = \frac{- 3 2 + 1 6}{- 4}$	$x = \frac{- 3 2 - 1 6}{- 4}$
$x = \frac{- 1 6}{- 4}$	$x = \frac{- 4 8}{- 4}$
$x = 4$	$x = 12$

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

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

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

x = \frac{- 1 6}{- 4}

x = \frac{- 4 8}{- 4}

x = 4

x = 12

Value of the Discriminant	Number of Real Solutions
$b^{2} - 4 a c > 0$	$2$
$b^{2} - 4 a c = 0$	$1$
$b^{2} - 4 a c < 0$	$0$

Value of the Discriminant

Number of Real Solutions

b^{2} - 4 a c > 0

2

b^{2} - 4 a c = 0

1

b^{2} - 4 a c < 0

0

Solve the equation using the Quadratic Formula. If needed, round to one decimal.

x^{2} - 12 x + 36 = 0

x^{2} + 7 x + 16 = 0

2 x^{2} - 6 x - 8 = 0

8 x^{2} + 8 = 6 - 9 x

We will use the Quadratic Formula to solve the quadratic equation. The formula relates to a quadratic equation in the following way. ax^2+ bx+ c=0 ⇓ x=- b± sqrt(b^2-4 a c)/2 a Let's identify the values of a, b, and c in the given equation. x^2-12x+36=0 ⇕ 1x^2+( - 12)x+ 36=0 We see that a= 1, b= - 12, and c= 36. Let's substitute these values into the Quadratic Formula and evaluate the right-hand side.

As we did in Part A, we first need to identify the values of a, b, and c in the standard form of the quadratic equation. x^2+7x+16=0 ⇓ 1x^2+ 7x+ 16=0 We see that a= 1, b= 7, and c= 16. If we substitute these values into the Quadratic Formula, we can calculate the solutions.

We cannot calculate the real square root of a negative number. Therefore, there are no real solutions to this equation.

Again, we begin by identifying the values of a, b, and c that are used in the Quadratic Formula. 2x^2-6x-8=0 ⇓ 2x^2+( -6)x+( -8)=0 Let's substitute the values into the Quadratic Formula and evaluate the right-hand side.

We found two solutions, x=4 and x=-1.

Before we can use the Quadratic Formula, we need to rewrite this equation in standard form.

Now we can identify the values of a, b, and c. 8x^2+ 9x+ 2=0 Let's substitute them into the Quadratic Formula and evaluate the right-hand side.

We found two solutions, x≈ -0.3 and x≈ -0.8.

Determine the number of real solutions of the following quadratic equations.

x^{2} - 6 x + 10 = 0

x^{2} - 5 x - 3 = 0

- 8 x^{2} + 8 x = 2

If we only want to know the number of solutions to a quadratic equation and not the solutions themselves, we can analyze the discriminant of the equation. Let's mark the discriminant in the Quadratic Formula. x=- b±sqrt(b^2-4ac)/2a We see that the quadratic equation is already given in its standard form, we can easily identify the values of a, b, and c. x^2 -6x+10=0 ⇓ 1x^2+( - 6)x+ 10=0 Now, let's evaluate the discriminant for the given values.

Since the discriminant is negative, the quadratic equation has no real solutions.

If the discriminant is greater than zero, the equation will have two real solutions. If it is equal to zero, the equation will have one real solution. Finally, if the discriminant is less than zero, the equation will have no real solutions.

Again, we only want to know the number of solutions, so we will consider the value of the discriminant. Let's first identify the values of a, b, and c of the standard form of the given quadratic equation. x^2 -5x-3=0 ⇓ 1x^2+( - 5)x+( - 3)=0 Now, let's evaluate the discriminant for these given values.

Since the discriminant is a positive number, the quadratic equation has two real solutions.

Just as in the previous parts, we only need to consider the value of the discriminant. To find this value, we start by identifying the values of a, b, and c. However, our actual first step is to rewrite the equation in its standard form. -8x^2+8x=2 ⇓ -8x^2+ 8x+( - 2)=0 Now we can evaluate the discriminant for these values.

Since the discriminant is zero, the quadratic equation has one real solution.

How many $x -$ intercepts does the graph of the function have?

y = x^{2} + 5 x - 1

y = 4 x^{2} + 4 x + 1

y = - 6 x^{2} + 3 x - 4

A function reaches its x-intercept when y=0. Therefore, to find how many solutions the quadratic function has, we must first set it equal to zero and then analyze the discriminant of the Quadratic Formula. Let's start by identifying the discriminant in the Quadratic Formula. x=- b±sqrt(b^2-4ac)/2a Next, let's set the function equal to zero and identify the values of a, b, and c. x^2 +5x-1=0 ⇓ 1x^2+ 5x+( -1)=0 Now we can evaluate the discriminant for the given values.

The discriminant is positive, which means the graph has two x-intercepts.

As we did in Part A, we will set the function equal to zero and then identify the values of a, b, and c. 4x^2 +4x+1=0 ⇓ 4x^2+ 4x+ 1=0 Now, let's evaluate the discriminant of the Quadratic Formula for these values.

The discriminant is zero. This means the graph of the function has one x-intercept.

As we did in the previous parts, we will set the function equal to zero and identify the values of a, b, and c. -6x^2+3x-4=0 ⇓ -6x^2+ 3x+( - 4)=0 Let's evaluate the discriminant of the quadratic function for these values.

The discriminant is negative, which means the graph of the function has no x-intercepts.

Davontay is throwing a ball over a

4

meter tall wall. The ball's trajectory can be described by the following model.

h (x) = 2.5 + 2 x - 0.5 x^{2}

In this model,

h

is the height of the ball in meters and

x

is the horizontal distance of the ball from Devontay, also in meters.

What is the domain of the function? Write the answer as a compound inequality.

What is the range of the function? Write the answer as a compound inequality.

How far away from the wall must Davontay stand in order for the ball to make it over? Choose the correct answer.

The domain of a function is the set of values that we can substitute for the independent variable x. Since x measures the horizontal distance of the wall from Davontay and a distance cannot be negative, we know that negative values of x are not possible. With this information, we can write the lower part of the domain. x≥ 0 Let's find the upper part of the domain. Notice that the ball cannot go below the ground. Therefore, we have an upper restriction to the domain where the function intercepts the positive part of the x-axis.

To find this value of x, we must solve the equation when h(x)=0. Notice that we are only interested in the positive x-intercept as we have already established that x must be greater than or equal to 0.

Let's identify the values of a, b, and c. - 0.5x^2+2x+2.5=0 ⇓ -0.5x^2+ 2x+ 2.5=0 We see that a= -0.5, b= 2, and c= 2.5. Let's substitute these values into the Quadratic Formula and evaluate the right-hand side.

The positive x-intercept is 5. Now we can write the domain. 0≤ x≤ 5

The range gives a function's possible output values. As we already explained in Part A, the ball cannot fall below the ground, which means that h(x) must be greater than or equal to 0. h(x)≥ 0 To find the upper limit of the range, we need to determine the function's greatest value. A parabola reaches its maximum value at the vertex. Let's find the x-coordinate of the vertex by determining the line of symmetry. This can be found by using the following formula. x_s=- b/2 a Recall that in Part A we also found that a= -0.5 and b= 2.

The axis of symmetry is given by the equation x_s=2.

To determine the greatest value of the function, or its vertex, we will substitute x=2 into the function rule and evaluate.

The function's greatest value is 4.5. Now we can write the range as below. 0≤ h ≤ 4.5

In order for the ball to make it over the wall, the ball's height must be greater than 4 meters when it reaches the wall. Let's set h(x) equal to 4 and solve for x.

Let's identify the values of a, b, and c for this new equation. - 0.5x^2+2x-1.5=0 ⇓ -0.5x^2+ 2x+( -1.5)=0 Now we can calculate the solutions to the new equation.

The ball reaches a height of 4 feet, the height of the wall, when it has traveled 1 and 3 meters away from Davontay. This means that it will hit the upper part of the wall and not make it over if he stands exactly this distance away. Therefore, when Davontay stands more than 1 meter but less than 3 meters from the wall, the ball will make it over.

This means that Davontay needs to stand more than 1 meter but less than 3 meters away from the wall in order for the ball to make it over.

Use the given information to find the length of the rectangle's longer side.

We have been given the area of the rectangle and also expressions describing its length and width. Because both length and width must be positive, start by determining values of variable x for which these measures are positive. x+2 > 0 &⇒ x > -2 2x+3 > 0 &⇒ x > - 32 Combining these two constraints gives us that x should be greater than - 32 for both measures to be positive. Next, recall that a rectangle's area is calculated by multiplying its length and width. A= l w ⇒ A=( 2x+3)( x+2) We know that the area is 91 square feet. Let's substitute 91 for A in the formula and proceed to write the quadratic equation in standard form.

Now that we have the equation in standard form, we can use the Quadratic Formula to solve for x. Notice that we should disregard any solution that is less than or equal to - 32 because this would make at least one of the side measures negative.

Using the Quadratic Formula, we found that the solutions of the given equation are x = -7 ± 274. Let's now evaluate both positive and negative case to check if the solutions are greater than - 32.

x=-7 ± 27/4
x=-7 + 27/4	x=-7 - 27/4
x=20/4	x=-34/4
x=5 ✓	x=-17/2 *

Now that we know that x=5, we can calculate the measurements of the sides of the rectangle by substituting x=5 into the expressions.

Side Length	Substitute	Evaluate
x+2	5+2	7
2x+3	2(5)+3	13

Therefore, the length of the longer side is 13 feet.

Solving Quadratic Equations With the Quadratic Formula

Catch-Up and Review

Is There a Solution?

The Quadratic Formula

Quadratic Formula

Proof

Solving a Quadratic Equation Using the Quadratic Formula

Hint

Solution

Solving a Quadratic Equation Not in Standard Form

Hint

Solution

Solving a Quadratic Equation Using the Quadratic Formula

Discriminant of a Quadratic Equation

Discriminant

Determining Whether There is a Solution Without Solving

Hint

Solution

Determining the Number of Real Solutions of a Quadratic Equation

Determining the Number of Solutions of a Quadratic Equation Without Solving

Hint

Solution

Solving Quadratic Equations With the Quadratic Formula

Recommended exercises

	10 Theory slides
	10 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions