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## 6.7: Applications Involving Quadratic Equations

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Learning Objectives

• Set up and solve applications involving relationships between real numbers.
• Set up and solve applications involving geometric relationships involving area and the Pythagorean theorem.
• Set up and solve applications involving the height of projectiles.

## Number Problems

The algebraic setups of the word problems that we have previously encountered led to linear equations. When we translate the applications to algebraic setups in this section, the setups lead to quadratic equations. Just as before, we want to avoid relying on the “guess and check” method for solving applications. Using algebra to solve problems simplifies the process and is more reliable.

Example $$\PageIndex{1}$$

One integer is $$4$$ less than twice another integer, and their product is $$96$$. Set up an algebraic equation and solve it to find the two integers.

First, identify the variables. Avoid two variables by using the relationship between the two unknowns.

The key phrase, “their product is $$96$$,” indicates that we should multiply and set the product equal to $$96$$.

$$n\cdot (2n-4)=96$$

Once we have the problem translated to a mathematical equation, we then solve. In this case, we can solve by factoring. The first step is to write the equation in standard form:

$$\begin{array} {cc} {n\cdot (2n-4)=96}&{\color{Cerulean}{Distribute\:n.}}\\{2n^{2}-4n=96}&{\color{Cerulean}{Subtract\:96\:from\:both\:sides.}}\\{2n^{2}-4n-96=0}&{} \end{array}$$

Next, factor completely and set each variable factor equal to zero.

$$\begin{array}{cc}{2n^{2}-4n-96=0}&{\color{Cerulean}{Factor\:out\:the\:GCF,\:2.}}\\{2(n^{2}-2n-48)=0}&{\color{Cerulean}{Factor\:the\:resulting\:trinomial.}}\\{2(n+6)(n-8)=0}&{\color{Cerulean}{Set\:each\:variable\:factor\:equal\:to\:zero.}} \end{array}$$

$$\begin{array}{ccc}{n+6=0}&{\text{or}}&{n-8=0}\\{n=-6}&{}&{n=8} \end{array}$$

The problem calls for two integers whose product is $$+96$$. The product of two positive numbers is positive and the product of two negative numbers is positive. Hence we can have two sets of solutions. Use $$2n−4$$ to determine the other integers.

$$\begin{array} {cc} {n=-6}&{n=8}\\{2n-4=2(\color{OliveGreen}{-6}\color{black}{)-4}}&{2n-4=2(\color{OliveGreen}{8}\color{black}{)-4}}\\{=-12-4}&{=16-4}\\{=-16}&{=12} \end{array}$$

Two sets of integers solve this problem: {$$8, 12$$} and {$$−6, −16$$}. Notice that $$(8)(12) = 96$$ and $$(−6)(−16) = 96$$; our solutions check out.

With quadratic equations, we often obtain two solutions for the identified unknown. Although it may be the case that both are solutions to the equation, they may not be solutions to the problem. If a solution does not solve the original application, then we disregard it.

Recall that consecutive odd and even integers both are separated by two units.

Example $$\PageIndex{2}$$

• The product of two consecutive positive odd integers is $$99$$. Find the integers.

Let $$n$$ represent the first positive odd integer.

Let $$\color{OliveGreen}{n+2}$$ represent the next positive odd integer.

The key phase, “product…is 99,” indicates that we should multiply and set the product equal to $$99$$.

$$n\cdot (n+2)=99$$

Rewrite the quadratic equation in standard form and solve by factoring.

\begin{aligned} n^{2}+2n&=99 \\ n^{2}+2n-99&=0 \\ (n-9)(n+11)&=0 \end{aligned}

$$\begin{array}{ccc}{n-9=0}&{\text{or}}&{n+11=0}\\{n=9}&{}&{n=-11} \end{array}$$

Because the problem asks for positive integers, $$n=9$$ is the only solution. Back substitute to determine the next odd integer.

\begin{aligned} n+2&=\color{OliveGreen}{9}\color{black}{+2} \\ &=11 \end{aligned}

The consecutive positive odd integers are $$9$$ and $$11$$.

Example $$\PageIndex{3}$$

Given two consecutive positive odd integers, the product of the larger and twice the smaller is equal to $$70$$. Find the integers.

Let $$n$$ represent the smaller positive odd integer.

Let $$n+2$$ represent the next positive odd integer.

The key phrase “twice the smaller” can be translated to $$2n$$. The phrase “product…is 70” indicates that we should multiply this by the larger odd integer and set the product equal to $$70$$.

$$(n+2)\cdot 2n=70$$

Solve by factoring.

$$\begin{array}{cc}{(n+2)\cdot 2n=70}&{\color{Cerulean}{Distribute.}}\\{2n^{2}+4n=70}&{\color{Cerulean}{Subtract\:70\:from\:both\:sides.}}\\{2n^{2}+4n-70=0}&{\color{Cerulean}{Factor\:out\:the\:GCF,\:2.}}\\{2(n^{2}+2n-35)=0}&{\color{Cerulean}{Factor\:the\:resulting\:trinomial.}}\\{2(n-5)(n+7)=0}&{\color{Cerulean}{Set\:each\:variable\:factor\:equal\:to\:zero.}} \end{array}$$

$$\begin{array}{ccc}{n-5=0}&{\text{or}}&{n+7=0}\\{n=5}&{}&{n=-7} \end{array}$$

Because the problem asks for positive integers, $$n=5$$ is the only solution.

Back substitute into $$n + 2$$ to determine the next odd integer.

\begin{aligned} n+2&=\color{OliveGreen}{5}\color{black}{+2}\\&=7 \end{aligned}

The positive odd integers are $$5$$ and $$7$$.

Exercise $$\PageIndex{1}$$

The product of two consecutive positive even integers is $$168$$. Find the integers.

The positive even integers are $$12$$ and $$14$$.

## Geometry Problems

When working with geometry problems, it is helpful to draw a picture. Below are some area formulas that you are expected to know. (Recall that $$π≈3.14$$.)

Example $$\PageIndex{4}$$

The floor of a rectangular room has a length that is $$4$$ feet more than twice its width. If the total area of the floor is $$240$$ square feet, then find the dimensions of the floor.

Use the formula $$A=l⋅w$$ and the fact that the area is $$240$$ square feet to set up an algebraic equation.

\begin{aligned} A&=l\cdot w \\ \color{OliveGreen}{240}&\color{black}{=(}\color{OliveGreen}{2w+4}\color{black}{)\cdot w} \end{aligned}

$$\begin{array} {ccc} {w-10=0}&{\text{or}}&{w+12=0} \\{w=10}&{}&{w=-12} \end{array}$$

At this point we have two possibilities for the width of the rectangle. However, since a negative width is not defined, choose the positive solution, $$w=10$$. Back substitute to find the length.

\begin{aligned} 2w+4&=2(\color{OliveGreen}{10}\color{black}{)+4} \\ &=20+4 \\ &=24 \end{aligned}

The width is $$10$$ feet and the length is $$24$$ feet.

It is important to include the correct units in the final presentation of the answer. In the previous example, it would not make much sense to say the width is $$10$$. Make sure to indicate that the width is $$10$$ feet.

Example $$\PageIndex{5}$$

The height of a triangle is $$3$$ inches less than twice the length of its base. If the total area of the triangle is $$7$$ square inches, then find the lengths of the base and height.

Use the formula $$A=\frac{1}{2}bh$$ and the fact that the area is $$7$$ square inches to set up an algebraic equation.

\begin{aligned} A&=\frac{1}{2} b\cdot h \\ \color{OliveGreen}{7}&\color{black}{=\frac{1}{2}b(\color{OliveGreen}{2b-3}\color{black}{)}} \end{aligned}

To avoid fractional coefficients, multiply both sides by $$2$$ and then rewrite the quadratic equation in standard form.

Factor and then set each factor equal to zero.

$$\begin{array}{ccc}{2b-7=0}&{\text{or}}&{b+2=0}\\{2b=7}&{}&{b=-2}\\{b=\frac{7}{2}}&{}&{} \end{array}$$

In this case, disregard the negative answer; the length of the base is $$\frac{7}{2}$$ inches long. Use $$2b−3$$ to determine the height of the triangle.

The base measures $$\frac{7}{2} = 3\frac{1}{2}$$ inches and the height is $$4$$ inches.

The base of a triangle is $$5$$ units less than twice the height. If the area is $$75$$ square units, then what is the length of the base and height?

The height is $$10$$ units and the base is $$15$$ units.

Recall that a right triangle is a triangle where one of the angles measures $$90$$°. The side opposite of the right angle is the longest side of the triangle and is called the hypotenuse. The Pythagorean theorem gives us a relationship between the legs and hypotenuse of any right triangle, where $$a$$ and $$b$$ are the lengths of the legs and $$c$$ is the length of the hypotenuse:

Given certain relationships, we use this theorem when determining the lengths of sides of right triangles.

Example $$\PageIndex{6}$$

The hypotenuse of a right triangle is $$10$$ inches. If the short leg is $$2$$ inches less than the long leg, then find the lengths of the legs.

Given that the hypotenuse measures $$10$$ inches, substitute its value into the Pythagorean theorem and obtain a quadratic equation in terms of $$x$$.

\begin{aligned} a^{2}+b^{2}&=c^{2} \\ (\color{OliveGreen}{x-2}\color{black}{)^{2}+}\color{OliveGreen}{x}\color{black}{^{2}}&=\color{OliveGreen}{10}\color{black}{^{2}} \end{aligned}

Multiply and rewrite the equation in standard form.

\begin{aligned} (x-2)^{2}+x^{2}&=10^{2} \\ x^{2}-4x+4+x^{2}&=100 \\ 2x^{2}-4x-96&=0 \end{aligned}

Once it is in standard form, factor and set each variable factor equal to zero.

\begin{aligned} 2x^{2}-4x-96&=0\\ 2(x^{2}-2x-48)&=0 \\ 2(x+6)(x-8)&=0 \end{aligned}

$$\begin{array}{ccc}{x+6=0}&{\text{or}}&{x-8=0}\\{x=-6}&{}&{x=8} \end{array}$$

Because lengths cannot be negative, disregard the negative answer. In this case, the long leg measures $$8$$ inches. Use $$x−2$$ to determine the length of the short leg.

\begin{aligned} x-2&=\color{OliveGreen}{8}\color{black}{-2} \\ &=6 \end{aligned}

The short leg measures $$6$$ inches and the long leg measures $$8$$ inches.

Example $$\PageIndex{7}$$

One leg of a right triangle measures $$3$$ centimeters. The hypotenuse of the right triangle measures $$3$$ centimeters less than twice the length of the unknown leg. Find the measure of all the sides of the triangle.

To set up an algebraic equation, we use the Pythagorean theorem.

\begin{aligned} a^{2}+b^{2}&=c^{2} \\ \color{OliveGreen}{3}\color{black}{^{2}+}\color{OliveGreen}{x}\color{black}{^{2}}&=(\color{OliveGreen}{2x-3}\color{black}{)^{2}} \end{aligned}

\begin{aligned} 3^{2}+x^{2}&=(2x-3)^{2} \\ 9+x^{2}&=4x^{2}-12x+9 \\ 0&=3x^{2}-12x \\ 0&=3x(x-4) \end{aligned}

$$\begin{array}{ccc}{3x=0}&{\text{or}}&{x-4=0}\\{x=0}&{}&{x=4} \end{array}$$

Disregard $$0$$. The length of the unknown leg is $$4$$ centimeters. Use $$2x−3$$ to determine the length of the hypotenuse.

The sides of the triangle measure $$3$$ centimeters, $$4$$ centimeters, and $$5$$ centimeters.

Exercise $$\PageIndex{2}$$

The hypotenuse of a right triangle measures $$13$$ units. If one leg is $$2$$ units more than twice that of the other, then find the length of each leg.

The two legs measure $$5$$ units and $$12$$ units.

## Projectile Problems

The height of an object launched upward, ignoring the effects of air resistance, can be modeled with the following formula:

$\text{height}=-\frac{1}{2}gt^{2}+v_{0}t+s_{0}$

Using function notation, which is more appropriate, we have

$h(t)=-\frac{1}{2}gt^{2}+v_{0}t+s_{0}$

With this formula, the height can be calculated at any given time $$t$$ after the object is launched. The coefficients represent the following:

We consider only problems where the acceleration due to gravity can be expressed as $$g=32$$ ft/sec$$^{2}$$. Therefore, in this section time will be measured in seconds and the height in feet. Certainly though, the formula is valid using units other than these.

Example $$\PageIndex{8}$$

The height of a projectile launched upward at a speed of $$32$$ feet/second from a height of $$128$$ feet is given by the function $$h(t)=−16t^{2}+32t+128$$. How long does it take to hit the ground?

An inefficient method for finding the time to hit the ground is to simply start guessing at times and evaluating. To do this, construct a chart.

Use the table to sketch the height of the projectile over time.

We see that at $$4$$ seconds, the projectile hits the ground. Note that when this occurs, the height is equal to $$0$$. Now we need to solve this problem algebraically. To find the solution algebraically, use the fact that the height is $$0$$ when the object hits the ground. We need to find the time, $$t$$, when $$h(t)=0$$.

$$h(t)=-16t^{2}+32t+128 \\ \color{Cerulean}{\downarrow} \\ 0 =-16t^{2}+32t+128$$

Solve the equation by factoring

\begin{aligned} 0 &=-16t^{2}+32t+128 \\ 0 &=-16(t^{2}-2t-8) \\ 0&=-16(t-4)(t+2) \end{aligned}

Now set each variable factor to zero.

$$\begin{array}{ccc}{t-4=0}&{\text{or}}&{t+2=0}\\{t=4}&{}&{t=-2} \end{array}$$

As expected, the projectile hits the ground at $$t=4$$ seconds. Disregard $$−2$$ as a solution because negative time is not defined.

The projectile hits the ground $$4$$ seconds after it is launched.

Example $$\PageIndex{9}$$

The height of a certain book dropped from the top of a $$144$$ foot building is given by $$h(t)=−16t^{2}+144$$. How long does it take to hit the ground?

Find the time $$t$$ when the height $$h(t)=0$$.

\begin{aligned}0&=-16t^{2}+144 \\ 0&=-16(t^{2}-9) \\ 0&=-16(t+3)(t-3) \end{aligned}

$$\begin{array}{ccc}{t+3=0}&{\text{or}}&{t-3=0}\\{t=-3}&{}&{t=3} \end{array}$$

The book takes $$3$$ seconds to hit the ground when dropped from the top of a $$144$$-foot building.

Exercise $$\PageIndex{3}$$

The height of a projectile, shot straight up into the air from the ground, is given by $$h(t)=−16t^{2}+80t$$. How long does it take to come back down to the ground?

It will take 5 seconds to come back down to the ground.

## Key Takeaways

• It is best to translate a word problem to a mathematical setup and then solve using algebra. Avoid using the “guess and check” method of solving applications in this section.
• When solving applications, check that your solutions make sense in the context of the question. For example, if you wish to find the length of the base of a triangle, then you would disregard any negative solutions.
• It is important to identify each variable and state in a sentence what each variable represents. It is often helpful to draw a picture.

Exercise $$\PageIndex{4}$$ Number Problems

Set up an algebraic equation and then solve.

• One integer is five times another. If the product of the two integers is $$80$$, then find the integers.
• One integer is four times another. If the product of the two integers is $$36$$, then find the integers.
• An integer is one more than four times another. If the product of the two integers is $$39$$, then find the integers.
• An integer is $$3$$ more than another. If the product of the two integers is $$130$$, then find the integers.
• An integer is $$2$$ less than twice another. If the product of the two integers is $$220$$, then find the integers.
• An integer is $$3$$ more than twice another. If the product of the two integers is $$90$$, then find the integers.
• One integer is $$2$$ units more than another. If the product of the two integers is equal to five times the larger, then find the two integers.
• A positive integer is $$1$$ less than twice another. If the product of the two integers is equal to fifteen times the smaller, then find the two integers.
• A positive integer is $$3$$ more than twice a smaller positive integer. If the product of the two integers is equal to six times the larger, then find the integers.
• One positive integer is $$3$$ more than another. If the product of the two integers is equal to twelve times the smaller, then find the integers.
• An integer is $$3$$ more than another. If the product of the two integers is equal to $$2$$ more than four times their sum, then find the integers.
• An integer is $$5$$ more than another. If the product of the two integers is equal to $$2$$ more than twice their sum, then find the integers.
• The product of two consecutive positive even integers is $$120$$. Find the integers.
• The product of two consecutive positive integers is $$110$$. Find the integers.
• The product of two consecutive positive integers is $$42$$. Find the integers.
• The product of two consecutive positive odd integers is equal to $$1$$ less than seven times the sum of the integers. Find the integers.
• The product of two consecutive positive even integers is equal to $$22$$ more than eleven times the sum of the integers. Find the integers.
• The sum of the squares of two consecutive positive odd integers is $$74$$. Find the integers.
• The sum of the squares of two consecutive positive even integers is $$100$$. Find the integers.
• The sum of the squares of two consecutive positive integers is $$265$$. Find the integers.
• The sum of the squares of two consecutive positive integers is $$181$$. Find the integers.
• For two consecutive positive odd integers, the product of twice the smaller and the larger is $$126$$. Find the integers.
• For two consecutive positive even integers, the product of the smaller and twice the larger is $$160$$. Find the integers.

1. {$$4, 20$$} or {$$−4, −20$$}

3. $$3, 13$$

5. {$$11, 20$$} or {$$−22, −10$$}

7. {$$5, 7$$} or {$$−2, 0$$}

9. $$6, 15$$

11. {$$7, 10$$} or {$$−2, 1$$}

13. $$10, 12$$

15. $$10, 11$$

17. $$13, 15$$

19. $$5, 7$$

21. $$11, 12$$

23. $$7, 9$$

Exercise $$\PageIndex{5}$$ Geometry Problems

• The width of a rectangle is $$7$$ feet less than its length. If the area of the rectangle is $$170$$ square feet, then find the length and width.
• The length of a rectangle is $$2$$ feet more than its width. If the area of the rectangle is $$48$$ square feet, then find the length and width.
• The width of a rectangle is $$3$$ units less than the length. If the area is $$70$$ square units, then find the dimensions of the rectangle.
• The width of a rectangle measures one half of the length. If the area is $$72$$ square feet, then find the dimensions of the rectangle.
• The length of a rectangle is twice that of its width. If the area of the rectangle is $$72$$ square inches, then find the length and width.
• The length of a rectangle is three times that of its width. If the area of the rectangle is $$75$$ square centimeters, then find the length and width.
• The length of a rectangle is $$2$$ inches more than its width. The area of the rectangle is equal to $$12$$ inches more than three times the perimeter. Find the length and width of the rectangle.
• The length of a rectangle is $$3$$ meters more than twice the width. The area of the rectangle is equal to $$10$$ meters less than three times the perimeter. Find the length and width of the rectangle.
• A uniform border is to be placed around an $$8$$-inch-by-$$10$$-inch picture. If the total area including the border must be $$224$$ square inches, then how wide should the border be?

10. A $$2$$-foot brick border is constructed around a square cement slab. If the total area, including the border, is $$121$$ square feet, then what are the dimensions of the slab?

11. The area of a picture frame including a $$2$$-inch wide border is $$99$$ square inches. If the width of the inner area is $$2$$ inches more than its length, then find the dimensions of the inner area.

12. A box can be made by cutting out the corners and folding up the edges of a square sheet of cardboard. A template for a cardboard box with a height of $$2$$ inches is given. What is the length of each side of the cardboard sheet if the volume of the box is to be $$50$$ cubic inches?

13. The height of a triangle is $$3$$ inches more than the length of its base. If the area of the triangle is $$44$$ square inches, then find the length of its base and height.

14. The height of a triangle is $$4$$ units less than the length of the base. If the area of the triangle is $$48$$ square units, then find the length of its base and height.

15. The base of a triangle is twice that of its height. If the area is $$36$$ square centimeters, then find the length of its base and height.

16. The height of a triangle is three times the length of its base. If the area is $$73\frac{1}{2}$$ square feet, then find the length of the base and height.

17. The height of a triangle is $$1$$ unit more than the length of its base. If the area is $$5$$ units more than four times the height, then find the length of the base and height of the triangle.

18. The base of a triangle is $$4$$ times that of its height. If the area is $$3$$ units more than five times the height, then find the length of the base and height of the triangle.

19. The diagonal of a rectangle measures $$5$$ inches. If the length is $$1$$ inch more than its width, then find the dimensions of the rectangle.

20. The diagonal of a rectangle measures $$10$$ inches. If the width is $$2$$ inches less than the length, then find the area of the rectangle.

21. If the sides of a right triangle are consecutive even integers, then what are their measures?

22. The hypotenuse of a right triangle is $$13$$ units. If the length of one leg is $$2$$ more than twice the other, then what are their lengths?

23. The shortest leg of a right triangle measures $$9$$ centimeters and the hypotenuse measures $$3$$ centimeters more than the longer leg. Find the length of the hypotenuse.

24. The long leg of a right triangle measures $$24$$ centimeters and the hypotenuse measures $$4$$ centimeters more three times the short leg. Find the length of the hypotenuse.

1. Length: $$17$$ feet; width: $$10$$ feet

3. Length: $$10$$ units; width: $$7$$ units

5. Length: $$12$$ inches; width: $$6$$ inches

7. Length: $$14$$ inches; width: $$12$$ inches

9. $$3$$ inches

11. $$5$$ inches by $$7$$ inches

13. Base: $$8$$ inches; height: $$11$$ inches

15. Base: $$12$$ centimeters; height: $$6$$ centimeters

17. Base: $$9$$ units; height: $$10$$ units

19. $$3$$ inches by $$4$$ inches

21. $$6$$ units, $$8$$ units, and $$10$$ units

23. $$15$$ centimeters

Exercise $$\PageIndex{6}$$ Projectile Problems

• The height of a projectile launched upward at a speed of $$32$$ feet/second from a height of $$48$$ feet is given by the function $$h(t)=−16t^{2}+32t+48. How long will it take the projectile to hit the ground? • The height of a projectile launched upward at a speed of \(16$$ feet/second from a height of $$192$$ feet is given by the function $$h(t)=−16t^{2}+16t+192$$. How long will it take to hit the ground?
• An object launched upward at a speed of $$64$$ feet/second from a height of $$80$$ feet. How long will it take the projectile to hit the ground?
• An object launched upward at a speed of $$128$$ feet/second from a height of $$144$$ feet. How long will it take the projectile to hit the ground?
• The height of an object dropped from the top of a $$64$$-foot building is given by $$h(t)=−16t^{2}+64$$. How long will it take the object to hit the ground?
• The height of an object dropped from an airplane at $$1,600$$ feet is given by $$h(t)=−16t^{2}+1,600$$. How long will it take the object to hit the ground?
• An object is dropped from a ladder at a height of $$16$$ feet. How long will it take to hit the ground?
• An object is dropped from a $$144$$-foot building. How long will it take to hit the ground?
• The height of a projectile, shot straight up into the air from the ground at $$128$$ feet/second, is given by $$h(t)=−16t^{2}+128t$$. How long does it take to come back down to the ground?
• A baseball, tossed up into the air from the ground at $$32$$ feet/second, is given by $$h(t)=−16t^{2}+32t$$. How long does it take to come back down to the ground?
• How long will it take a baseball thrown into the air at $$48$$ feet/second to come back down to the ground?
• A football is kicked up into the air at $$80$$ feet/second. Calculate how long will it hang in the air.

1. $$3$$ seconds

3. $$5$$ seconds

5. $$2$$ seconds

7. $$1$$ second

9. $$8$$ seconds

11. $$3$$ seconds

Exercise $$\PageIndex{7}$$ Discussion Board Topics

• Research and discuss the life of Pythagoras.
• If the sides of a square are doubled, then by what factor is the area increased? Why?
• Design your own geometry problem involving the area of a rectangle or triangle. Post the question and a complete solution on the discussion board.
• Write down your strategy for setting up and solving word problems. Share your strategy on the discussion board.

## Mathematics 9 Quarter 1-Module 9: Solving Problems Involving Quadratic Equations (Week 4 Learning Code – M9AL-1e-1)

In the previous lessons, you learned about the different ways in solving quadratic equations, the concepts of the “nature of its roots” and the relationship of its roots and coefficients. Your mastery of the lessons is an important tool to solve many real-world problems related to quadratic equations. In this module, you will learn how to solve word problems that involves quadratic equations.

LEARNING COMPETENCY

The learners will be able to:

• solve problems involving quadratic equations. M9AL-1e-1

## Can't Find What You'RE Looking For?

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## 10.7 Quadratic Word Problems: Age and Numbers

Quadratic-based word problems are the third type of word problems covered in MATQ 1099, with the first being linear equations of one variable and the second linear equations of two or more variables. Quadratic equations can be used in the same types of word problems as you encountered before, except that, in working through the given data, you will end up constructing a quadratic equation. To find the solution, you will be required to either factor the quadratic equation or use substitution.

Example 10.7.1

The sum of two numbers is 18, and the product of these two numbers is 56. What are the numbers?

First, we know two things:

$\begin{array}{l} \text{smaller }(S)+\text{larger }(L)=18\Rightarrow L=18-S \\ \\ S\times L=56 \end{array}$

Substituting $18-S$ for $L$ in the second equation gives:

$S(18-S)=56$

Multiplying this out gives:

$18S-S^2=56$

Which rearranges to:

$S^2-18S+56=0$

Second, factor this quadratic to get our solution:

$\begin{array}{rrrrrrl} S^2&-&18S&+&56&=&0 \\ (S&-&4)(S&-&14)&=&0 \\ \\ &&&&S&=&4, 14 \end{array}$

$\begin{array}{l} S=4, L=18-4=14 \\ \\ S=14, L=18-14=4 \text{ (this solution is rejected)} \end{array}$

Example 10.7.2

The difference of the squares of two consecutive even integers is 68. What are these numbers?

The variables used for two consecutive integers (either odd or even) is $x$ and $x + 2$. The equation to use for this problem is $(x + 2)^2 - (x)^2 = 68$. Simplifying this yields:

$\begin{array}{rrrrrrrrr} &&(x&+&2)^2&-&(x)^2&=&68 \\ x^2&+&4x&+&4&-&x^2&=&68 \\ &&&&4x&+&4&=&68 \\ &&&&&-&4&&-4 \\ \hline &&&&&&\dfrac{4x}{4}&=&\dfrac{64}{4} \\ \\ &&&&&&x&=&16 \end{array}$

This means that the two integers are 16 and 18.

Example 10.7.3

The product of the ages of Sally and Joey now is 175 more than the product of their ages 5 years prior. If Sally is 20 years older than Joey, what are their current ages?

The equations are:

$\begin{array}{rrl} (S)(J)&=&175+(S-5)(J-5) \\ S&=&J+20 \end{array}$

Substituting for S gives us:

$\begin{array}{rrrrrrrrcrr} (J&+&20)(J)&=&175&+&(J&+&20-5)(J&-&5) \\ J^2&+&20J&=&175&+&(J&+&15)(J&-&5) \\ J^2&+&20J&=&175&+&J^2&+&10J&-&75 \\ -J^2&-&10J&&&-&J^2&-&10J&& \\ \hline &&\dfrac{10J}{10}&=&\dfrac{100}{10} &&&&&& \\ \\ &&J&=&10 &&&&&& \end{array}$

This means that Joey is 10 years old and Sally is 30 years old.

For Questions 1 to 12, write and solve the equation describing the relationship.

• The sum of two numbers is 22, and the product of these two numbers is 120. What are the numbers?
• The difference of two numbers is 4, and the product of these two numbers is 140. What are the numbers?
• The difference of two numbers is 8, and the sum of the squares of these two numbers are 320. What are the numbers?
• The sum of the squares of two consecutive even integers is 244. What are these numbers?
• The difference of the squares of two consecutive even integers is 60. What are these numbers?
• The sum of the squares of two consecutive even integers is 452. What are these numbers?
• Find three consecutive even integers such that the product of the first two is 38 more than the third integer.
• Find three consecutive odd integers such that the product of the first two is 52 more than the third integer.
• The product of the ages of Alan and Terry is 80 more than the product of their ages 4 years prior. If Alan is 4 years older than Terry, what are their current ages?
• The product of the ages of Cally and Katy is 130 less than the product of their ages in 5 years. If Cally is 3 years older than Katy, what are their current ages?
• The product of the ages of James and Susan in 5 years is 230 more than the product of their ages today. What are their ages if James is one year older than Susan?
• The product of the ages (in days) of two newborn babies Simran and Jessie in two days will be 48 more than the product of their ages today. How old are the babies if Jessie is 2 days older than Simran?

Example 10.7.4

Doug went to a conference in a city 120 km away. On the way back, due to road construction, he had to drive 10 km/h slower, which resulted in the return trip taking 2 hours longer. How fast did he drive on the way to the conference?

The first equation is $r(t) = 120$, which means that $r = \dfrac{120}{t}$ or $t = \dfrac{120}{r}$.

For the second equation, $r$ is 10 km/h slower and $t$ is 2 hours longer. This means the second equation is $(r - 10)(t + 2) = 120$.

We will eliminate the variable $t$ in the second equation by substitution:

$(r-10)(\dfrac{120}{r}+2)=120$

Multiply both sides by $r$ to eliminate the fraction, which leaves us with:

$(r-10)(120+2r)=120r$

Multiplying everything out gives us:

$\begin{array}{rrrrrrrrr} 120r&+&2r^2&-&1200&-&20r&=&120r \\ &&2r^2&+&100r&-&1200&=&120r \\ &&&-&120r&&&&-120r \\ \hline &&2r^2&-&20r&-&1200&=&0 \end{array}$

This equation can be reduced by a common factor of 2, which leaves us with:

$\begin{array}{rrl} r^2-10r-600&=&0 \\ (r-30)(r+20)&=&0 \\ r&=&30\text{ km/h or }-20\text{ km/h (reject)} \end{array}$

Example 10.7.5

Mark rows downstream for 30 km, then turns around and returns to his original location. The total trip took 8 hr. If the current flows at 2 km/h, how fast would Mark row in still water?

If we let $t =$ the time to row downstream, then the time to return is $8\text{ h}- t$.

The first equation is $(r + 2)t = 30$. The stream speeds up the boat, which means $t = \dfrac{30}{(r + 2)}$, and the second equation is $(r - 2)(8 - t) = 30$ when the stream slows down the boat.

We will eliminate the variable $t$ in the second equation by substituting $t=\dfrac{30}{(r+2)}$:

$(r-2)\left(8-\dfrac{30}{(r+2)}\right)=30$

Multiply both sides by $(r + 2)$ to eliminate the fraction, which leaves us with:

$(r-2)(8(r+2)-30)=30(r+2)$

$\begin{array}{rrrrrrrrrrr} (r&-&2)(8r&+&16&-&30)&=&30r&+&60 \\ &&(r&-&2)(8r&+&(-14))&=&30r&+&60 \\ 8r^2&-&14r&-&16r&+&28&=&30r&+&60 \\ &&8r^2&-&30r&+&28&=&30r&+&60 \\ &&&-&30r&-&60&&-30r&-&60 \\ \hline &&8r^2&-&60r&-&32&=&0&& \end{array}$

This equation can be reduced by a common factor of 4, which will leave us:

$\begin{array}{rll} 2r^2-15r-8&=&0 \\ (2r+1)(r-8)&=&0 \\ r&=&-\dfrac{1}{2}\text{ km/h (reject) or }r=8\text{ km/h} \end{array}$

For Questions 13 to 20, write and solve the equation describing the relationship.

• A train travelled 240 km at a certain speed. When the engine was replaced by an improved model, the speed was increased by 20 km/hr and the travel time for the trip was decreased by 1 hr. What was the rate of each engine?
• Mr. Jones visits his grandmother, who lives 100 km away, on a regular basis. Recently, a new freeway has opened up, and although the freeway route is 120 km, he can drive 20 km/h faster on average and takes 30 minutes less time to make the trip. What is Mr. Jones’s rate on both the old route and on the freeway?
• If a cyclist had travelled 5 km/h faster, she would have needed 1.5 hr less time to travel 150 km. Find the speed of the cyclist.
• By going 15 km per hr faster, a transit bus would have required 1 hr less to travel 180 km. What was the average speed of this bus?
• A cyclist rides to a cabin 72 km away up the valley and then returns in 9 hr. His speed returning is 12 km/h faster than his speed in going. Find his speed both going and returning.
• A cyclist made a trip of 120 km and then returned in 7 hr. Returning, the rate increased 10 km/h. Find the speed of this cyclist travelling each way.
• The distance between two bus stations is 240 km. If the speed of a bus increases by 36 km/h, the trip would take 1.5 hour less. What is the usual speed of the bus?
• A pilot flew at a constant speed for 600 km. Returning the next day, the pilot flew against a headwind of 50 km/h to return to his starting point. If the plane was in the air for a total of 7 hours, what was the average speed of this plane?

Example 10.7.6

Find the length and width of a rectangle whose length is 5 cm longer than its width and whose area is 50 cm 2 .

First, the area of this rectangle is given by $L\times W$, meaning that, for this rectangle, $L\times W=50$, or $(W+5)W=50$.

Multiplying this out gives us:

$W^2+5W=50$

$W^2+5W-50=0$

Second, we factor this quadratic to get our solution:

$\begin{array}{rrrrrrl} W^2&+&5W&-&50&=&0 \\ (W&-&5)(W&+&10)&=&0 \\ &&&&W&=&5, -10 \\ \end{array}$

We reject the solution $W = -10$.

This means that $L = W + 5 = 5+5= 10$.

Example 10.7.7

If the length of each side of a square is increased by 6, the area is multiplied by 16. Find the length of one side of the original square.

The relationship between these two is:

$\begin{array}{rrl} \text{larger area}&=&16\text{ times the smaller area} \\ (x+12)^2&=&16(x)^2 \end{array}$

Simplifying this yields:

$\begin{array}{rrrrrrr} x^2&+&24x&+&144&=&16x^2 \\ -16x^2&&&&&&-16x^2 \\ \hline -15x^2&+&24x&+&144&=&0 \end{array}$

Since this is a problem that requires factoring, it is easiest to use the quadratic equation:

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a},\hspace{0.25in}\text{ where }a=-15, b=24\text{ and }c=144$

Substituting these values in yields $x = 4$ or $x=-2.4$ (reject).

Example 10.7.8

Nick and Chloe want to surround their 60 by 80 cm wedding photo with matting of equal width. The resulting photo and matting is to be covered by a 1 m 2 sheet of expensive archival glass. Find the width of the matting.

$(L+2x)(W+2x)=1\text{ m}^2$

$(80\text{ cm }+2x)(60\text{ cm }+2x)=10,000\text{ cm}^2$

$4800+280x+4x^2=10,000$

$4x^2+280x-5200=0$

Which reduces to:

$x^2 + 70x - 1300 = 0$

Second, we factor this quadratic to get our solution.

It is easiest to use the quadratic equation to find our solutions.

$x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a},\hspace{0.25in}\text{ where }a=1, b=70\text{ and }c=-1300$

Substituting the values in yields:

$x=\dfrac{-70\pm \sqrt{70^2-4(1)(-1300)}}{2(1)}\hspace{0.5in}x=\dfrac{-70\pm 10\sqrt{101}}{2}$

$x=-35+5\sqrt{101}\hspace{0.75in} x=-35-5\sqrt{101}\text{ (rejected)}$

For Questions 21 to 28, write and solve the equation describing the relationship.

• Find the length and width of a rectangle whose length is 4 cm longer than its width and whose area is 60 cm 2 .
• Find the length and width of a rectangle whose width is 10 cm shorter than its length and whose area is 200 cm 2 .
• A large rectangular garden in a park is 120 m wide and 150 m long. A contractor is called in to add a brick walkway to surround this garden. If the area of the walkway is 2800 m 2 , how wide is the walkway?
• A park swimming pool is 10 m wide and 25 m long. A pool cover is purchased to cover the pool, overlapping all 4 sides by the same width. If the covered area outside the pool is 74 m 2 , how wide is the overlap area?
• In a landscape plan, a rectangular flowerbed is designed to be 4 m longer than it is wide. If 60 m 2 are needed for the plants in the bed, what should the dimensions of the rectangular bed be?
• If the side of a square is increased by 5 units, the area is increased by 4 square units. Find the length of the sides of the original square.
• A rectangular lot is 20 m longer than it is wide and its area is 2400 m 2 . Find the dimensions of the lot.
• The length of a room is 8 m greater than its width. If both the length and the width are increased by 2 m, the area increases by 60 m 2 . Find the dimensions of the room.

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1. Solving quadratic equations COLLECTION (Bundle)

2. Using Quadratic Equations to Solve Problems

3. solving problems using quadratic models

4. Using Quadratic Equations to Solve Problems

5. PROBLEM SOLVING INVOLVING QUADRATIC EQUATIONS

6. Solving A Quadratic Equation By Factoring

#### VIDEO

1. 10th Maths

2. Solving Problems Involving Quadratic Equations

3. Solving Problems Involving Quadratic Equations

4. 10th Maths

6. Solving Problems Involving Quadratic Equations (Part 2)

1. PDF Lesson 13: Application Problems with Quadratic Equations

Solve the problems below. Write the problem, your work, and the solution in the text box below to submit your work. Be sure to show all of your work. Here is a link explaining how to show your work. We suggest saving your work in a word processor. Solve each problem below showing the steps as indicated in the lesson. 1.

2. PDF 7.4 Solving Problems Using Quadratic Equations

Just like with Quadratic Functions, there are two main types of applications associated with Quadratic Equations. Type I, the equation is provided and for Type II, the object is to come up with the equation yourself. Type I - Equation Provided Example 1: A golf ball is hit from the top of a tower that is 24 m high. The ball follows a ...

Quadratic Equations. mc-TY-quadeqns-1. This unit is about the solution of quadratic equations. These take the form ax2 +bx+c = 0. We will look at four methods: solution by factorisation, solution by completing the square, solution using a formula, and solution using graphs. In order to master the techniques explained here it is vital that you ...

4. PDF Solve each equation with the quadratic formula.

Create your own worksheets like this one with Infinite Algebra 1. Free trial available at KutaSoftware.com. ©x d2Q0D1S2L RKcuptra2 GSRoYfRtDwWa8r9eb NLOL1Cs.j 4 lA0ll x TrCiagFhYtKsz OrVe4s4eTrTvXeZdy.c I RM8awd7e6 ywYiPtghR OItnLfpiqnAiutDeY QALlegpe6bSrIay V1g.N.

5. PDF CHAPTER 8 Quadratic Equations, Functions, and Inequalities

This quantity divided by the quantity 2a is the Quadratic Formula. 124. The Quadratic Formula is derived by solving the general quadratic equation ax2 bx. c 0 by the method of completing the square. > 0 opens upward. 70. 106. use the method of completing the square to write the function in standard form f x a x h 2 k.

6. PDF GCSE TOPIC BOOKLET PROBLEMS INVOLVING QUADRATIC EQUATIONS

In the triangle below, show that x satisfies the equation 3x2 — o. -3x cm [3] 300 Solve the equation 3x2 — place. — 2) cm Diagram not drawn to scale — O and hence find the length of BC, correct to 1 decimal [4]

7. 9.6: Solve Applications of Quadratic Equations

Step 5: Solve the equation. Substitute in the values. Distribute. This is a quadratic equation; rewrite it in standard form. Solve the equation using the Quadratic Formula. Identify the $$a,b,c$$ values. Write the Quadratic Formula. Then substitute in the values of $$a,b,c$$. Simplify. Figure 9.5.13: Rewrite to show two solutions.

Method 1: "Solve for x2". Although the x2 is not "by itself," we can use the multiplication principle to write an equivalent equation in the special form to apply the basic strategy. 16x2 = 25 16x2 = 25 16 16 x2 = 25 16. The resulting equivalent equation is in the special form to apply our basic strat-. egy.

A quadratic equation takes the form ax2 +bx+c =0 where a, b and c are numbers. The number a cannot be zero. In this unit we will look at how to solve quadratic equations using four methods: • solution by factorisation • solution by completing the square • solution using a formula • solution using graphs Factorisation and use of the ...

Step 1: This equation is in standard form. But we want the terms that contain the variable to be on the left and the constant to be on the right. So we add 6 to both sides, obtaining 2− =6 The equation is now in the proper form for completing the square. Step 2: Because b (the coefficient of x) is -1, 𝑏 2 is −1 2 and (𝑏 2) 2 is (−1 2) 2

Steps: Add or subtract terms so that one side of the equation equals 0. Factor the polynomial expression. Set each factor equals to 0 and solve for the unknown. Remark: if two of the factors are the same, then the solution is said to be a double root or a root of multiplicity two. Solve the following equation.

12. PDF QUADRATIC FUNCTIONS, PARABOLAS, AND PROBLEM SOLVING

2.5 Quadratic Functions, Parabolas, and Problem Solving 99 Graphs of quadratic functions For the quadratic functionf~x! 5 ax2 1 bx 1 c: The graph is a parabola with axis of symmetry x 5 2b 2a. The parabola opensupward if a . 0, downward if a , 0. To ﬁnd the coordinates of the vertex,set x 5 2b 2a.Thenthey-coordinate is given by y 5 fS 2b 2a D.

13. PDF QUADRATIC EQUATIONS WORD PROBLEMS

To work out the problem we can define the sides of the triangle ac cording to the figure below: Step 1 - Write the equation x 2 + (x + 3) 2 = (x + 6) 2 Step 2 - Solve the equation By using the SQUARE OF A BINOMIAL FORMULA x 2 + x 2 + 6 x + 9 = x 2 + 12 x + 36 2x 2 + 6 x + 9 = x 2 + 12 x + 36 x 2 − 6x − 27 = 0 (x − 9)( x + 3) = 0 x − 9 = 0

14. PDF 9.7 Solve Systems with Quadratic Equations

techniques to solve a system of equations involving nonlinear equations, such as quadratic equations. Recall that the substitution method consists of the following three steps. STEP 1 Solve one of the equations for one of its variables. STEP 2 Substitute the expression from Step 1 into the other equation and solve for the other variable.

15. PDF 9 Solving Quadratic Equations

Now you will use square roots to solve quadratic equations of the form ax2 c 0. First isolate x2 = on one side of the equation to obtain x2 d. Then solve by taking the square root = of each side. • When d 0, x2 d has one real solution, x 0. 3x2 27 0 by factoring. • When d < 0, x2 d has no real solutions.

16. 6.7: Applications Involving Quadratic Equations

Number Problems. The algebraic setups of the word problems that we have previously encountered led to linear equations. When we translate the applications to algebraic setups in this section, the setups lead to quadratic equations. Just as before, we want to avoid relying on the "guess and check" method for solving applications.

17. Mathematics 9 Quarter 1-Module 9: Solving Problems Involving Quadratic

Your mastery of the lessons is an important tool to solve many real-world problems related to quadratic equations. In this module, you will learn how to solve word problems that involves quadratic equations. LEARNING COMPETENCY. The learners will be able to: solve problems involving quadratic equations. M9AL-1e-1; Math9_Quarter1_Module9_FINAL-V3-1

18. Q1w5-Math 9 Module 1

30 sri is eics e es 10 es 5153 Equations of the form ax2 + bx + c = 0 To solve x2 - 4x + 3 = 0 we need to factorise the algebraic expression and then use the Null Factor Law. If p × q = 0, then at least one of p and q must be zero. x2 - 4x + 3 = 0 (x - 3)(x - 1) = 0So either x - 3 = 0 or x - 1 = 0 ∴ x = 3 or x = 1 Substituting these values into the original equation will show that they are ...

20. 10.7 Quadratic Word Problems: Age and Numbers

Quadratic-based word problems are the third type of word problems covered in MATQ 1099, with the first being linear equations of one variable and the second linear equations of two or more variables. Quadratic equations can be used in the same types of word problems as you encountered before, except that, in working through the given data, you ...

21. Solving Quadratic and Rational Equations

Math9_Q2-Mod3 - Free download as PDF File (.pdf), Text File (.txt) or read online for free. This document is a mathematics module that covers solving problems involving quadratic equations and rational algebraic equations. It includes two lessons that teach students how to solve word problems involving these types of equations. The module provides learning objectives, icons to guide students ...

22. Solving Problems Involving Quadratic Equations

10. Solving Problems Involving Quadratic Equations - Free download as Word Doc (.doc / .docx), PDF File (.pdf), Text File (.txt) or read online for free. This document outlines a lesson plan on solving quadratic equations. The lesson begins with motivating students on the importance of solving quadratic equations to model real-world problems.