Browse Tag: quadratic word problems

Solving Quadratic Word Problems in Algebra

Quadratic Equations are equations of the form ax^2 + bx + c = 0 where a, b and c are real numbers and a \neq 0. Depending on the form of the equation, you can solve for x by extracting the quare root, factoring, or using the quadratic formula.This type of equation appears in various problems that involves multiplication and usually appears in the Civil Service Exams.

The following series details the method and strategies in solving problems involving quadratic equations.

How to Solve Quadratic Word Problems Part 1 is about solving problems involving consecutive integers. In this problem, the product of consecutive numbers is given and factoring was used to solve the problem.  Continue Reading

How to Solve Quadratic Word Problems Part 3

In the previous two posts (Part 1 and Part 2), we have discussed two problems involving quadratic equations. The first one is

A rectangular flower garden with dimensions 3m by 7m is surrounded by a walk of uniform width. If the area of the walk is 11 square meters, what is the width of the walk in meters?


If we let x be the width of the walk, then the length of the walk becomes x + 7 + x = 7 + 2x and the width becomes x + 3 + x = 3 + 2x as shown in the figure below.


We know that the area of the garden (the inner rectangle) is equal 7 \times 3 = 21 square meters. The area of the larger rectangle is equal to

(3 + 2x)(7 + 2x) = 21 + 14x + 6x + 4x^2 = 21 + 20x + 4x^2.

Now, to get the area of the walk, we have to subtract the area of the garden from the area of the large rectangle. That is,

area of the larger rectangle – area of the smaller rectangle = 11

4x^2 + 20x + 21 - 21 = 11

4x^2 + 20x = 11

Subtracting 11 from both sides, we have

4x^2 + 20x -11 = 0.

Now, the quadratic equation is not factorable, so we use the quadratic formula in order to find the value of x. In using the quadratic formula, we want to identify the values of a, b, and c in the equation ax^2 + bx + c = 0 and substitute in the quadratic formula

x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

From the equation above, a = 4, b = 20 and c = -11. Substituting these values in the quadratic formula, we have

x = \dfrac{-20 \pm \sqrt{(-20)^2 - 4(4)(-11)}}{2(4)}

= \dfrac{-20 \pm \sqrt{400 + 176}}{8}

= \dfrac{20 \pm 24}{8}

That means that we have two roots

x_1 = \dfrac{-20 + 24}{8} = \dfrac{1}{2}

x_2 = \dfrac{-20 - 24}{8} = \dfrac{-44}{8} = \dfrac{-11}{2}.

As we can see, we x_2 is not a possible root because it is negative. Therefore, the acceptable answer to the problem is

x_1 = \dfrac{1}{2}.

This means that the width of the walk is ½ meters.

Check: If the width of the walk is ½ meters, then the length of the larger rectangle is 1/2 + 7 + 1/2 = 8 and the width is 1/2 + 3 + 1/2 = 4. The area is (8)(4) = 32 sq. m.

Now the area of the walk is 21 meters and 32 - 21 = 11 sq. m. which is the area of the walk indicated in the problem. Therefore, we are correct.

You can also watch the Youtube video below regarding the discussion above.