Chapter 10: Quadratics 10.7 Quadratic Word Problems: Age and NumbersQuadratic-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. 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:
Substituting
Multiplying this out gives:
Which rearranges to:
Second, factor this quadratic to get our solution:
Therefore:
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
This means that the two integers are 16 and 18. 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:
Substituting for S gives us:
This means that Joey is 10 years old and Sally is 30 years old. QuestionsFor Questions 1 to 12, write and solve the equation describing the relationship.
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 For the second equation, We will eliminate the variable
Multiply both sides by
Multiplying everything out gives us:
This equation can be reduced by a common factor of 2, which leaves us with:
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 The first equation is We will eliminate the variable
Multiply both sides by
Multiplying everything out gives us:
This equation can be reduced by a common factor of 4, which will leave us:
QuestionsFor Questions 13 to 20, write and solve the equation describing the relationship.
Find the length and width of a rectangle whose length is 5 cm longer than its width and whose area is 50 cm2. First, the area of this rectangle is given by Multiplying this out gives us:
Which rearranges to:
Second, we factor this quadratic to get our solution:
We reject the solution This means that 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:
Simplifying this yields:
Since this is a problem that requires factoring, it is easiest to use the quadratic equation: Substituting these values in yields 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 m2 sheet of expensive archival glass. Find the width of the matting.
Or, in cm:
Multiplying this out gives us:
Which rearranges to:
Which reduces to:
Second, we factor this quadratic to get our solution. It is easiest to use the quadratic equation to find our solutions. Substituting the values in yields: QuestionsFor Questions 21 to 28, write and solve the equation describing the relationship.
<a class=”internal” href=”/intermediatealgebraberg/back-matter/answer-key-10-7/”>Answer Key 10.7 |