topic 7.3.2

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Topic 7.3.2Topic 7.3.2

Applications of QuadraticsApplications of Quadratics

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Topic7.3.2


California Standards:20.0 Students use the quadratic formula to find the roots of a second-degree polynomial and to solve quadratic equations.

What it means for you:You’ll model real-life problems using quadratic equations and then solve them using the quadratic formula.

Key words:• quadratic formula• completing the square

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Topic7.3.2


Sometimes you will have to make a “mathematical model” first, and then solve the equations you get from it.

When you get your solutions, you have to interpret them.

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Topic7.3.2

Modeling Means Writing Your Own Equations


Example 1 shows how to model a real-life situation as a quadratic equation.

To solve some math problems you need to use quadratic equations — but the question won’t always give you the equations.

You need to be able to write your own equations from the information you’re given — this is called modeling.

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Topic7.3.2

Example 1

Solution follows…


Solution continues…

Solution

The difference between a pair of numbers is 9. Find all such pairs of numbers that have a product of 220.

There are two different numbers in each pair — call the lower number x. Then the higher number is x + 9.

Write the information from the question in the form of an equation: x(x + 9) = 220

This is a quadratic equation, so rearrange it to the form ax2 + bx + c = 0.

x(x + 9) = 220 x2 + 9x = 220 x2 + 9x – 220 = 0

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Topic7.3.2

Example 1


Solution (continued)


Write down a, b, and c: a = 1, b = 9, and c = –220.

Now you can use the quadratic formula.

or

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Topic7.3.2

Example 1




So there are two possible values for x (where x is the lower of the two numbers): x = –20 or x = 11.

The higher of the two numbers is found by adding 9 to each of these values. So there are two possible pairs of numbers, and they are: –20 and –11 and 11 and 20.

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Topic7.3.2

Example 2

Solution follows…


Solution continues…

Solution

Find the dimensions of the rectangle whose length is 7 in. more than twice its width, and whose area is 120 in2.

Let x = width in inches. Then the length is 2x + 7 inches.

From the question: x(2x + 7) = 120

Rearrange this quadratic into standard form:

2x2 + 7x – 120 = 02x2 + 7x = 120x(2x + 7) = 120

Now use the formula with a = 2, b = 7, and c = –120.

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Topic7.3.2

Example 2




So x = 6.19 or x = –9.69.

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Topic7.3.2

Example 2




x = 6.19 or x = –9.69. But the width cannot be negative, so you can ignore x = –9.69.

So the width must be x = 6.2 in. (to 1 decimal place), and the length is then 2x + 7 = (2 × 6.19) + 7 = 19.4 in. (to 1 decimal place).

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Topic7.3.2

Guided Practice

Solution follows…


1. The difference between two numbers is 7. Find all possible pairs of such numbers if the product of the two numbers is 198.

2. Find the dimensions of a rectangular garden whose length is 10 meters more than three times its width, if the area is 77 m2.

Let the smaller number = x, then the other number is x + 7.From the question, x(x + 7) = 198, so x2 + 7x – 198 = 0.Using the quadratic equation, x = 11 or x = –18.So the numbers are 11 and 18 or –18 and –11.

Let the width = w, then the length is 3w + 10.From the question, w(3w + 10) = 77, so 3w2 + 10w – 77 = 0.Using the quadratic equation, w = 3.67 or x = –7.Ignore the negative value, so the width is 3.67 m and the length is 21 m.

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Topic7.3.2

Guided Practice

Solution follows…


3. Twice the square of a number is equal to eight times the number. Find the number.

4. The sum of the squares of two consecutive odd integers is 74. Find the numbers.

Let the number = x.From the question, 2x2 = 8x, so 2x2 – 8x = 0.Using the quadratic equation, x = 0 or x = 4.So the number is 0 or 4.

Let the numbers be x and x + 2.From the question, x2 + (x + 2)2 = 74, so 2x2 + 4x – 70 = 0.Using the quadratic equation, x = –7 or x = 5.So the numbers are –7 and –5 or 5 and 7.

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Topic7.3.2

Guided Practice

Solution follows…


5. The sum of the squares of two consecutive even integers is 340. Find the possible numbers.

6. The length of a rectangular field is 10 meters less than four times its width. Find the dimensions if its area is 750 square meters.

Let the numbers be x and x + 2.From the question, x2 + (x + 2)2 = 340, so 2x2 + 4x – 336 = 0.Using the quadratic equation, x = 12 or x = –14.So the numbers are 12 and 14 or –14 and –12.

Let the width = w, then the length is 4w – 10.From the question, w(4w – 10) = 750, so 4w2 – 10w – 750 = 0.Using the quadratic equation, w = 15 or x = –12.5.Ignore the negative value, so the width is 15 m and the length is 50 m.

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Topic7.3.2

Guided Practice

Solution follows…


7. When 15 and 19 are each increased by t, the product of the resulting numbers is 837. Find the value(s) of t.

8. A mother is three times as old as her daughter. Four years ago the product of their ages was 256. Find their current ages.

The numbers are (15 + t) and (19 + t).From the question, (15 + t )(19 + t) = 837, so t2 + 34t – 552 = 0.Using the quadratic equation, t = 12 or t = –46.

Let the daughter’s current age = d, then the mother’s current age 3d.Four years ago, their ages were d – 4 and 3d – 4.From the question, (d – 4)(3d – 4) = 256, so 3d2 – 16d – 240 = 0.Using the quadratic equation, d = 12 or d = –3.3.Ignore the negative value, so the daughter is 12 and her mother is 36.

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Topic7.3.2

Independent Practice

Solution follows…


1. A man is five times older than his son. In three years’ time, the product of their ages will be 380. Find their ages now.

2. Lorraine is 10 years older than Ahanu. In three years’ time the product of their ages will be 600. Find Ahanu and Lorraine’s ages now.

7 and 35 years old

17 and 27 years old

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Topic7.3.2


Solution follows…


3. A picture of 10 inches by 7 inches is in a frame whose area (including the space for the picture) is 154 square inches. Find the dimensions of the frame if the gap between the edge of the picture and the frame is the same all the way around.

14 in by 11 in

10 in7 inx x

x

x

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Topic7.3.2


Solution follows…


4. Jennifer has a picture of her boyfriend Zach measuring 10 inches by 8 inches. She frames the picture in a frame that has an area of 224 square inches (including the space for the picture). Find the dimensions of the picture frame if the gap between the edge of the picture and the frame is the same all the way around.

5. A wire of length 50 feet is bent to form a rectangular figure that has no overlap. If the area of the figure formed is 144 square feet, find the dimensions of the figure.

16 in by 14 in

9 ft by 16 ft

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Topic7.3.2


Solution follows…


6. A piece of wire 22 yards long is bent to form a rectangular figure whose area is 28 square yards. Find the dimensions of the figure, given that there is no overlap in the wire.

7. Show that the sum of the solutions of 4x2 – 4x – 3 = 0

is equal to 1 (= – ).

8. Show that the product of the solutions of x2 – 7x + 10 = 0

is equal to 10 (= ).

4 yd by 7 yd

4x2 – 4x – 3 = (2x – 3)(2x + 1)

So x = or x = – . The sum is + – = 13

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1

2

3

2

1

2

x2 – 7x + 10 = (x – 5)(x – 2)

So x = 5 or x = 2. The product is 5 × 2 = 10.

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Topic7.3.2


Solution follows…


9. Using the quadratic formula, show that the sum of the solutions of the general quadratic equation ax2 + bx + c = 0

is equal to – , and that the product of the roots is .

The solutions are and .

The product is

The sum of these is:

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Topic7.3.2


Round UpRound Up

Quadratic equations pop up a lot in Algebra I.

If you know the quadratic formula then you’ll always be able to solve them by just substituting the values into the formula.

topic 7.3.2

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