8. Solving Quadratic Equations With the Quadratic Formula
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Chapter 8
8.
Solving Quadratic Equations With the Quadratic Formula
This lesson delves into the Quadratic Formula, a powerful tool for solving quadratic equations. The formula offers a systematic approach to find the solutions of any quadratic equation written in standard form. The method is presented, proven, and applied to various scenarios. One real-world example involves Magdalena and Diego, who face a challenge while camping. They need to determine if a stone thrown by Diego can reach a branch 15 feet above the ground. Using the Quadratic Formula, they can predict the stone's trajectory and decide if it will hit the target. Another example showcases Magdalena's fundraiser for paralympic athletes. She aims to determine the ticket price to ensure a profit of at least $200. The Quadratic Formula helps her find the optimal ticket price. The lesson also highlights the discriminant, a part of the formula, which indicates the number of real solutions a quadratic equation has. Overall, the Quadratic Formula is not just a mathematical concept but a practical tool with diverse applications.
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Lesson Settings & Tools
| | 10 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Solving Quadratic Equations With the Quadratic Formula
Slide of 10
There are several methods for solving quadratic equations. In this lesson, the Quadratic Formula will be presented, proven, and used. Additionally, a method for determining the number of solutions without actually solving the equation will be presented.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Try a few practice exercises as a warm-up!
a Simplify the numeric expression sqrt(4^2 - 4(1)(2)) by using the properties of square roots to remove the perfect-square factor.
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b Consider the quadratic function y = (x+2)^2-1. Of the following expressions, which represents the same function written in standard form?
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c Identify the coefficient b of the quadratic equation x-5=2(x-3)^2 when written in standard form ax^2+bx+c = 0.
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Is There a Solution?
Magdalena and Diego, both huge fans of statistics, went camping to bond under the stars and talk stats. However, they realize that bears are in the area. They need to hang their food basket from a branch 15 feet above the ground. Diego figures he can throw a stone with a rope attached to it over the branch. As Diego winds up, Magdalena sheepishly snickers, "No way that works."
In wondering if Diego's throw will be a success, consider the following quadratic function that models the height, in air, of the stone's location after t seconds of being thrown. h(t) = 5(-3t^2 + 5t + 1)
Magdalena also wonders what quadratic equation represents this scenario. Help her find it. Then, without solving the equation, determine whether it is even possible to know if the stone will reach the branch.Discussion
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The Quadratic Formula
Besides graphing, using square roots, factoring, and completing the square, there is another method for solving a quadratic equation. This method consists of using the Quadratic Formula. Check out how to derive the formula by completing the square!
Rule
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Quadratic Formula
The Quadratic Formula can be used to solve a quadratic equation written in standard form ax^2+bx+c = 0.
x=- b±sqrt(b^2-4ac)/2a
Proof
The Quadratic Formula can be derived by completing the square given the standard form of the quadratic equation ax^2 + bx + c = 0. This method will be used to isolate the x-variable. To complete the square, there are five steps to follow.
1
Factor Out the Coefficient of x^2
expand_more It is easier to complete the square when the quadratic expression is written in the form x^2 + bx + c. Therefore, the coefficient a should be factored out. ax^2+bx+c = 0 ⇕ a(x^2+b/ax+c/a) = 0 Since the equation is quadratic, the coefficient a is not equal to 0. Therefore, both sides of the equation can be divided by a. a(x^2+b/ax+c/a) = 0 ⇕ x^2+b/ax+c/a = 0
2
Identify the Constant Needed to Complete the Square
expand_more The next step is to rewrite the equation by moving the existing constant to the right-hand side. To do so, ca will be subtracted from both sides of the equation. x^2+b/ax+ c/a = 0 ⇕ x^2+ b/ax = - c/a
The constant needed to complete the square can now be identified by focusing on the x-term, while ignoring the rest. One way to find this constant is by squaring half the coefficient of the x-term, which in this case is ba.
Note that leaving the constant as a power makes the next steps easier to perform.
3
Complete the Square
expand_more The square can now be completed by adding the constant found in Step 2 to both sides of the equation. \begin{gathered} x^2+\dfrac{b}{a}x = \text{-} \dfrac{c}{a} \\ \Updownarrow \\ \underbrace{x^2+\dfrac{b}{a}x + {\color{#FD9000}{\left(\dfrac{b}{2a} \right)^2}}}_\text{Perfect Square Trinomial} = \underbrace{\text{-} \dfrac{c}{a} + {\color{#FD9000}{\left(\dfrac{b}{2a} \right)^2}}}_\text{Constant} \end{gathered} The first three terms form a perfect square trinomial, which can be factored as the square of a binomial. The other two terms do not contain the variable x. Therefore, their value is constant.
4
Factor the Perfect Square Trinomial
expand_more The perfect square trinomial can now be factored and rewritten as the square of a binomial.
The process of completing the square is now finished.
x^2+b/ax + (b/2a )^2 = - c/a + (b/2a )^2
DenomMultFracToNumber
2 * a/2= a
x^2+2(b/2a)x + (b/2a )^2 = - c/a + (b/2a )^2
CommutativePropMult
Commutative Property of Multiplication
x^2+2x(b/2a) + (b/2a )^2 = - c/a + (b/2a )^2
FacPosPerfectSquare
a^2+2ab+b^2=(a+b)^2
(x + b/2a)^2 = - c/a + (b/2a )^2
5
Simplify the Equation
expand_more Finally, the right-hand side of the equation can be simplified.
Now, there is only one x-term. To isolate x, it is necessary to take square roots on both sides of the equation. This results in both a positive and a negative term on the right-hand side.
sqrt((x + b/2a)^2) = sqrt(b^2-4ac/4a^2) ⇕ x + b/2a = ± sqrt(b^2-4ac/4a^2)
Now, the equation can be further simplified to isolate x.
Finally, the Quadratic Formula has been obtained.
(x + b/2a)^2 = - c/a + (b/2a )^2
▼
Simplify right-hand side
CommutativePropAdd
Commutative Property of Addition
(x + b/2a)^2 = (b/2a )^2 - c/a
PowQuot
(a/b)^m=a^m/b^m
(x + b/2a)^2 = b^2/(2a)^2 - c/a
PowProdII
(a b)^m=a^m b^m
(x + b/2a)^2 = b^2/4a^2 - c/a
ExpandFrac
a/b=a * 4a/b * 4a
(x + b/2a)^2 = b^2/4a^2 - c* 4a/a * 4a
CommutativePropMult
Commutative Property of Multiplication
(x + b/2a)^2 = b^2/4a^2 - 4ac/4a* a
ProdToPowTwoFac
a* a=a^2
(x + b/2a)^2 = b^2/4a^2 - 4ac/4a^2
SubFrac
Subtract fractions
(x + b/2a)^2 = b^2-4ac/4a^2
x + b/2a = ± sqrt(b^2-4ac/4a^2)
▼
Solve for x
SqrtQuot
sqrt(a/b)=sqrt(a)/sqrt(b)
x + b/2a = ± sqrt(b^2-4ac)/sqrt(4a^2)
SqrtProd
sqrt(a* b)=sqrt(a)*sqrt(b)
x + b/2a = ± sqrt(b^2-4ac)/2sqrt(a^2)
SqrtPowToNumber
sqrt(a^2)=a
x + b/2a = ± sqrt(b^2-4ac)/2a
SubEqn
LHS-b/2a=RHS-b/2a
x = - b/2a ± sqrt(b^2-4ac)/2a
MoveNegFracToNum
Put minus sign in numerator
x = - b/2a ± sqrt(b^2-4ac)/2a
AddSubFrac
Add and subtract fractions
x = - b ± sqrt(b^2-4ac)/2a
x = - b ± sqrt(b^2-4ac)/2a
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Solving a Quadratic Equation Using the Quadratic Formula
Magdalena will sell lottery tickets as a fundraiser to support paralympic athletes. The total profit p(x) depends on the price x of a ticket and can be modeled by using the following quadratic equation.
p(x) = -2x^2 + 32x + 104
Magdalena wants to raise at least $200. However, she has not yet set the price of each lottery ticket. Help Magdalena find the smallest amount that can be charged per ticket and still make a profit of at least $200. Round the price to the nearest whole dollar (the dollar sign is not necessary).
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Hint
Since the profit should be at least $200, let p(x) be equal to 200. Then, rewrite the quadratic equation in standard form. The equation can be solved using the Quadratic Formula.
Solution
It is given that the profit for the fundraiser should be at least $200. Consider the given quadratic equation, which models the profit, and substitute 200 for p(x). Then, rewrite the obtained equation in standard form.
Now, all of the coefficients in the standard form ax^2+bx+c = 0 can be determined.
-2x^2 + 32x - 96 = 0 ⇕ -2x^2 + 32x + ( - 96) = 0
Therefore, a= -2, b= 32, and c= -96. The obtained equation will be solved using the Quadratic Formula.
x = - b ± sqrt(b^2-4ac)/2a
The values of a, b, and c will now be substituted into the formula. Find x by evaluating the right-hand side of the formula.
Using the Quadratic Formula, it was obtained that the solutions for the equation are x = -32 ± 16-4. Finally, both solutions can be evaluated using a table.
p(x) = -2x^2 + 32x + 104
Substitute
p(x)= 200
200 = -2x^2 + 32x + 104
SubEqn
LHS-200=RHS-200
0 = -2x^2 + 32x - 96
RearrangeEqn
Rearrange equation
-2x^2 + 32x - 96 = 0
x = - b ± sqrt(b^2-4ac)/2a
SubstituteValues
Substitute values
x = - 32 ± sqrt(32^2-4( -2)( -96))/2( -2)
▼
Evaluate right-hand side
CalcPow
Calculate power
x=-32±sqrt(1024-4(-2)(-96))/2(-2)
MultPosNeg
a(- b)=- a * b
x=-32±sqrt(1024-(-8)(-96))/-4
MultNegNegTwoPar
(- a)(- b)=a* b
x=-32±sqrt(1024-768)/-4
SubTerm
Subtract term
x=-32±sqrt(256)/-4
CalcRoot
Calculate root
x = -32 ± 16/-4
| x = -32 ± 16/-4 | |
|---|---|
| x = -32 + 16/-4 | x = -32 - 16/-4 |
| x=-16/-4 | x=-48/-4 |
| x=4 | x=12 |
Since Magdalena wants the tickets to be as cheap as possible while making a profit of at least $200, the price each ticket should be $4.
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Solving a Quadratic Equation Not in Standard Form
A fire nozzle attached to a hose is a device used by firefighters to extinguish fires. Consider a firefighter who is aiming water to extinguish a fire on the third floor of a building. The base of the fire is situated 22 feet above the ground.
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Hint
What is the height of the water stream's peak? Write a quadratic equation and solve it using the Quadratic Formula.
Solution
Consider the given situation on a coordinate plane and assume that the firefighter is standing on the y-axis. 
Since the water stream's peak is 2 feet above the fire's base, whose height is 22 feet, its height h(x) is 2+22= 24 feet. This height will now be substituted into the equation of the given quadratic function to calculate the desired distance.
h(x) = -0.008x(x-100) + 4 ↓ 24 = -0.008x(x-100) + 4
The obtained quadratic equation can be solved using the Quadratic Formula. To do so, the equation must first be rewritten in standard form.
Next, the coefficients a, b, and c can be identified.
-0.008x^2 + 0.8x - 20 = 0 ⇕ -0.008x^2 + 0.8x + (- 20) = 0
Finally, these values will be substituted into the Quadratic Formula to solve the equation for x.
It has been found that this equation has exactly one solution, x = 50. Therefore, it can be said that the firefighter is standing at a horizontal distance of 50 feet from the water stream's peak.
24 = -0.008x(x-100) + 4
▼
Rewrite
Distr
Distribute -0.008x
24 = -0.008x^2+0.8x + 4
SubEqn
LHS-24=RHS-24
0 = -0.008x^2+0.8x -20
RearrangeEqn
Rearrange equation
-0.008x^2 + 0.8x - 20 = 0
x=- b±sqrt(b^2-4ac)/2a
SubstituteValues
Substitute values
x=- 0.8±sqrt(0.8^2-4( -0.008)(- 20))/2( -0.008)
▼
Evaluate right-hand side
CalcPow
Calculate power
x=-0.8±sqrt(0.64-4(-0.008)(-20))/2(-0.008)
MultPosNeg
a(- b)=- a * b
x=-0.8±sqrt(0.64-(-0.032)(-20))/-0.016
MultNegNegTwoPar
(- a)(- b)=a* b
x=-0.8±sqrt(0.64-0.64)/-0.016
SubTerm
Subtract term
x=-0.8±sqrt(0)/-0.016
CalcRoot
Calculate root
x = -0.8 ± 0/-0.016
AddSubTerms
Add and subtract terms
x = -0.8/-0.016
DivNegNeg
- a/- b=a/b
x = 0.8/0.016
CalcQuot
Calculate quotient
x = 50
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Solving a Quadratic Equation Using the Quadratic Formula
Solve the quadratic equations by using the Quadratic Formula. If necessary, round the answer to 2 decimal places.
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Discriminant of a Quadratic Equation
In general, quadratic equations have two, one or no real solutions. Before solving a quadratic equation, the number of real solutions can be determined by using the discriminant.
Concept
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Discriminant
In the Quadratic Formula, the expression b^2 - 4ac, which is under the radical symbol, is called the discriminant.
x = - b ± sqrt(b^2-4ac)/2a
A quadratic equation can have two, one, or no real solutions. Since the discriminant is under the radical symbol, its value determines the number of real solutions of a quadratic equation.
| Value of the Discriminant | Number of Real Solutions |
|---|---|
| b^2-4ac > 0 | 2 |
| b^2-4ac = 0 | 1 |
| b^2-4ac < 0 | 0 |
Moreover, the discriminant determines the number of x-intercepts of the graph of the related quadratic function.
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Determining Whether There is a Solution Without Solving
A farmer wants to build a fence around a vegetable garden. To make it simple, the farmer will build it in the shape of a rectangle. The farmer has enough wood to build a fence the length of 800 feet, including the gate.
The area of the vegetable garden changes as the side lengths of the rectangle change. The farmer wants the garden's area to be at least 50 000 square feet. Will he need to buy more wood to achieve this goal?
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Hint
Let x denote the length of one side of the rectangle. Then, use the fact that the length of the fence represents the perimeter of the rectangle. All things considered, how can the area of the rectangle be calculated?
Solution
Let x and y be the side lengths of the rectangle that represents the vegetable garden. The length of the fence is the perimeter P of the rectangle, which is the sum of all four side lengths.
P=x+x+y+y ⇕ P = 2x + 2y
It is given that the perimeter of the rectangle — how much wood the farmer has — is 800 feet. By substituting 800 for P, the equation can be solved for one of the side lengths, such as y.
The side lengths can now be placed in the diagram. It can be arbitrarily assumed that the length of the horizontal side is x. Keep in mind that both x and 400-x are measured in feet.
Next, the area of the rectangle will be calculated in terms of x. The area A of a rectangle is the product of the rectangle's length and width.
A = x(400-x)
The obtained formula for A is represented by a quadratic function. It is given that the farmer's desired area should be at least 50 000 square feet. Therefore, this number will be substituted for A in the formula.
A = x(400-x) ↓ 50 000 = x(400-x)
The above is a quadratic equation that is not written in standard form. Hence, the equation will be rewritten to determine the number of solutions. Determining the number of solutions will help find if a value for x exists so that the area of the rectangle is 50 000 square feet.
The equation is now in standard form. This means that the number of solutions can be determined using the discriminant. Next, the coefficients a, b, and c need to be identified. - x^2 + 400x - 50 000 = 0 ⇕ -1x^2 + 400x + (- 50 000) = 0
The variables a, b, and c can then be substituted into the discriminant b^2-4ac.
Since the discriminant is less than 0, there are no solutions to the equation. Therefore, the farmer will not be able to build a fence so that the area of the vegetable garden is 50 000 square feet. This means that he will need to buy more wood. Good thing he did the math before starting to construct the fence.
P = 2x+2y
Substitute
P= 800
800 = 2x+2y
▼
Solve for y
SubEqn
LHS-2x=RHS-2x
800 - 2x = 2y
DivEqn
.LHS /2.=.RHS /2.
800-2x/2 = y
WriteDiffFrac
Write as a difference of fractions
800/2 - 2x/2 = y
MovePartNumRight
a* b/c=a/c* b
800/2 - 2/2x = y
CalcQuot
Calculate quotient
400 - 1x = y
IdPropMult
Identity Property of Multiplication
400 - x = y
RearrangeEqn
Rearrange equation
y = 400 - x
50 000 = x(400-x)
▼
Rewrite
Distr
Distribute x
50 000 = 400x - x^2
SubEqn
LHS-50 000=RHS-50 000
0 = 400x - x^2 - 50 000
CommutativePropAdd
Commutative Property of Addition
0 = - x^2 + 400x - 50 000
RearrangeEqn
Rearrange equation
- x^2 + 400x - 50 000 = 0
b^2 - 4ac
SubstituteValues
Substitute values
400^2 - 4( -1)(- 50 000)
▼
Evaluate
CalcPow
Calculate power
160 000 - 4(-1)(-50 000)
MultPosNeg
a(- b)=- a * b
160 000 - (- 4)(-50 000)
MultNegNegOnePar
- a(- b)=a* b
160 000 - 200 000
SubTerm
Subtract term
-40 000
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Determining the Number of Real Solutions of a Quadratic Equation
Without solving the quadratic equations, use the discriminant to determine the number of real solutions.
Closure
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Determining the Number of Solutions of a Quadratic Equation Without Solving
The challenge presented at the beginning of this lesson asked if the stone thrown by Diego will reach, over some point in time, a branch located 15 feet above the ground.
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Hint
Substitute 15 for h(t) and identify the discriminant of the resulting quadratic equation.
Solution
Here, t represents the number of seconds that passed since Diego threw the stone. Since the branch is situated 15 feet above the ground, this number can be substituted for h(t) into the formula.
h(t) = 5(-3t^2 + 5t + 1) ↓ 15 = 5(-3t^2 + 5t + 1)
Although this quadratic equation can be solved using the Quadratic Formula, it is sufficient to find whether the equation has any solutions. To do so, the discriminant can be calculated. To calculate the discriminant, the equation needs to be rewritten in standard form.
Now, identify the coefficients a, b, and c in the obtained equation. -15t^2 + 25t - 10 = 0 ⇕ -15t^2 + 25t + ( -10) = 0
Finally, the values of the coefficients will be substituted into the discriminant.
Because the value of the discriminant is greater than 0, there are two solutions of the equation. Therefore, Diego and Magdalena know that the stone will reach the desired branch at two points in time.
15 = 5(-3t^2 + 5t + 1)
▼
Rewrite
Distr
Distribute 5
15 = -15t^2 + 25t + 5
SubEqn
LHS-15=RHS-15
0 = -15t^2 + 25t - 10
RearrangeEqn
Rearrange equation
-15t^2 + 25t - 10 = 0
b^2-4ac
SubstituteValues
Substitute values
25^2 - 4( -15)( -10)
▼
Evaluate
CalcPow
Calculate power
625 - 4(-15)(-10)
MultPosNeg
a(- b)=- a * b
625 - (- 60)(- 10)
MultNegNegOnePar
- a(- b)=a* b
625-600
SubTerm
Subtract term
25
Solving Quadratic Equations With the Quadratic Formula
Exercise 3.1
Enrique sells ice cream on a beach in Mexico. He buys the ice cream at 4 pesos per serving and sells it again for p pesos. Enrique theorizes that the total number of ice cream cones he can sell in a day x varies directly according to the following formula. x=310-12p Assume that Enrique sells every serving of ice cream he buys.
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Enrique's profit from sales is his revenue R less his costs C. We can write this as the following function. V(p) = R(p) - C(p) We can describe the profit Enrique makes from his sales by finding an expression for the revenue and costs with respect to p. Keep in mind that Enrique sells every serving of ice cream he buys.
- Revenue R(p): Enrique sells ice cream for p pesos per serving. If Enrique sells x servings of ice cream, he receives a revenue of px pesos in total.
- Costs C(p): Enrique buys a single ice cream for 4 pesos. If he purchases x ice cream servings, his costs are 4x pesos.
Because we have expressions for Enrique's costs and revenue, we can substitute these for R(p) and C(p) in our profit function. V(p) = px - 4x To write the function with respect to p only, we will eliminate x-variable by substituting x= 310-12p into the function rule and simplifying.
We wrote the function for Enrique's profit from sales in Part A. Notice that this is a quadratic function with a negative leading coefficient.
V(p) = - 12p^2+358p-1240
Therefore, the function reaches its maximum value at the vertex. The x-coordinate of the vertex is given by the equation of the axis of symmetry. We will write its equation by averaging the function's zeros. First we need to find them!
The profit is zero if Enrique sells the ice cream for 4 pesos or 1556 pesos. Next, we can calculate the mean of these values, p_1+p_22, to write the equation for the axis of symmetry of the parabola.
Enrique should sell his ice cream for about 15 pesos per serving to maximize his profit.
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Solving Quadratic Equations With the Quadratic Formula
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