Classwork Series and Exercises {Mathematics- SS3}: Word Problem Involving Quadratic Equations

Word Problem Involving Quadratic Equations

Write a “let” statement and equation for each example. Explain the difference.

1. Find two consecutive integers whose sum is 13

Let x = first integer

x + 1 = second integer

x + x + 1 = 13

Find two consecutive two consecutive integers whose products is 42

Let x = first integer

  X + 1 = second integer

 x(x + 1) = 42

Write “let” statement (if consecutive integers, use x; x + 1; x + 2 if consecutive even or odd, use x; x+2;x+4)

Write equation using keywords from statement: sum (add), product (multiply, more than (add), less than (subtract), etc.

Solve equation:

Check in original problem statement

1. Let x = first integer, x+2= second integer

x(x + 2) = 143

x² + 2x= 143

x² + 2x-143 = 0

(x + 13)(x – 11) = 0

Integers: -13 and -11 i.e. 11 and 13

Find two numbers whose sum is 9 and whose product is 20.

Solution

Let one of the numbers be x

The other number is 9-x

Their product is 20 : x(9-x) = 20

x(9 – x) = 20

i.e. 9x – x² – 20 = 0

Þ x² + 9x + 20 = 0 (the resulting quadratic equation)

Factorizing: (x – 4)(x – 5) = 0

\x = 4 or 5 √

\ The numbers are 4 and 5

If check sum of numbers: 4 + 5 = 9

The product of numbers:4 x 5 = 20

Example

The difference of two numbers is 2 and their product is 224. Find the numbers.

Let x and y be the numbers. Their difference is 2, so I can write

x – y = 2

Their product is 224, so

xy = 224

From x – y = 2, I get x = y + 2. Plug this into xy = 224 and solve for y:

(y + 2)y =224

y² + 2y =224

y² + 2y – 224 = 0

(y + 16)(y – 14) =224

If y = -16, then x = – 16 + 2 = – 14.

If y = 14, then x = 14 + 2 = 16.

So two pairs work: -14 and -16, and 14 and 16

Problem 

Motorboat moving upstream and downstream on a river. A motorboat makes a round trip on a river 56 miles upstream and 56 miles downstream, maintaining the constant speed 15 miles per hour relative to the water.

The entire trip up and back takes 7.5 hours.
What is the speed of the current?

Solution
Denote the unknown current speed of the river as x miles/hour.
When motorboat moves upstream, its speed relative to the bank of the river is 15 – x miles/hour, and the time spent moving upstream is 56/15 – x hours.
When motorboat moves downstream, its speed relative to the bank of the river is 15 + x miles/hour, and the time spent moving downstream is 56/15 + x hours.
So, the total time up and back is 56/15 – x + 56/15 + x, and it is equal to 7.5 hours, according to the problem input.
This gives an equation 56/15 – x + 56/15 + x = 7.5.
To simplify the equation, multiply both sides by (15 – x)(15 + x) and collect common terms. Step by step, you get 56(15 + x) + 56(15 – x) = 7.5(15 – x)(15 + x),
,1680 = 7.5(15² – x²)
1680/7.5 = 225 – x²

224 = 225 – x²

x² = 1.

Problem

 Andrew and Bill, working together, can cover the roof of a house in 6 days. Andrew, working alone, can complete this job in 5 days less than Bill. How long will it take Bill to make this job?

Solution
Denote x number of days for Bill to cover the roof, working himself, if Andrew works alone, he can complete this job in x – 5 days.
Thus, in one single day Andrew covers 1/x -5 part of the roof area, while Bill covers 1/x part of the roof area.
Working together, Andrew and Bill make 1/x – 5 ₊ 1/x of the whole work in each single day.
Since they can cover the entire roof in 6 days working together, the equation for the unknown value x is as follows: 6/x – 5 + 6/x = 1.

To simplify this equation, multiply both sides by (x – 5)x , then transfer all terms from the right side to the left with the opposite signs, then collect common terms and adjust the signs. In this way you get 6x + 6(x – 5) = x(x-5),

6x – 6x – 30 = x² – 5x,

-x² + 6x + 6x + 5x – 30 = 0,

– x² – 17x – 30 = 0,

You get the quadratic equation. Apply the quadratic formula to solve this equation. You get
x = 17 ± Ö17² – 4.30/2 = 17 ± Ö289 – 120/2 = 17 ± Ö169/2

Note: square root represent Ö

The equation has two roots: x₁ = 17 + 13/2 and x₂ = 17 -13/2 = 2 and.
The second root x₂ = 2 does not fit the given conditions (if Bill covers the roof in two days, then Andrew has 2 – 5= -3 days, what has no sense).

So, the potentially correct solution is x₁ = 15: Bill covers the roof in 15 days.
Let us check it. If Bill gets the job done in 15 days, then Andrew makes it in 10 days, working separately.
Since 6/10 + 6/15 = 1, this solution is correct.

Answer: Bill covers the roof in 15 days

EXERCISES

Lets see how much you’ve learnt, attach the following answers to the comment below

1. One leg of a right triangle exceeds the other leg by four inches.  The hypotenuse is 20 inches.  Find the length of the shorter leg of the right triangle. Hint: Pythagorean Theorem. A. 12 B. 14  16  D. 182.
2. The product of two consecutive integers is 56. Find the integers. Hint: Pythagorean Theorem A. -8, 7 -8, -7  C. 8, -7  D. -8, -8
3. The area of a rectangle is 80cm². If the length is 2cm more than the width, find the width. A. 9cm B 10cm C. 6cm D. 8cm
4. Solve this quadratic equation x² – 7x + 12.25 = 0, using Quadratic formula.
5. Solve these equations by factoring x: y = x² – 5x + 7  A. 1 and 7 B. 2 and 6 C. 1 and 6 D. 2 and 7

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