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?

Solution

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.

cs2

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}
and

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.

Enjoy!

Solving Quadratic Equations by Quadratic Formula

In the previous post, we have learned how to solve quadratic equations by factoring. In this post, we are going to learn how to solve quadratic equations using the quadratic formula. In doing this, we must identify the values of a, b, and c, in ax^2 + bx + c = 0 and substitute their values to the quadratic formula

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

Note that the value of a is the number in the term containing x^2, b is the number in the term containing x, and c is the value of the constant (without x or x^2).

The results in this calculation which are the values of x are the roots of the quadratic equation. Before you calculate using this formula, it is important that you master properties of radical numbers and how to calculate using them.

Example 1: Find the roots of x^2 - 4x - 4 = 0

Solution

From the equation, we can identify a = 1, b = -4, and c = -4.

Substituting these values in the quadratic formula, we have

x = \dfrac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-4)}}{2(1)}
x = \dfrac{4 \pm \sqrt{16 + 16}}{2}
x = \dfrac{4 \pm \sqrt{32}}{2}.

We know, that \sqrt{32} = \sqrt{(16)(2)} = \sqrt{16} \sqrt{2} = 4 \sqrt{2}. So, we have

x = \dfrac{4 \pm 4 \sqrt{2}}{2} = \dfrac{2(2 + \sqrt{2}}{2} = 2 \pm 2 \sqrt{2}

Therefore, we have two roots

2 + 2 \sqrt{2} or 2 - 2 \sqrt{2}

Example 2: Find the roots of 2x^2 - 6x = 15

Solution

Recall, that it easier to identify the values of a, b, and c if the quadratic equation is in the general form which is ax^2 + bx + c = 0. In order to make the right hand side of the equation above equal to 0, subtract 15 from both sides of the equation by 15. This results to

2x^2 - 6x - 15.

As we can see, a = 2, b = -6 and c =-15.

Substituting these values to the quadratic formula, we have

x = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4(2)(-15)}}{2(2)}

x = \dfrac{6 \pm \sqrt{36 + 120}}{4}

x = \dfrac{6 \pm \sqrt{156}}{4}.

But \sqrt{156} = \sqrt{(4)(39)} = \sqrt{4} \sqrt{39} = 2 \sqrt{39}.

Therefore,

x = \dfrac{6 \pm 2 \sqrt{39}}{4}.

Factoring out 2, we have

\dfrac{2(3 \pm \sqrt{39}}{4} = \dfrac{3 \pm \sqrt{39}}{2}

Therefore, we have two roots:

\dfrac{3 + \sqrt{39}}{2} or \dfrac{3 - \sqrt{39}}{2}

That’s it. In the next post, we are going to learn how to use quadratic equations on how to solve word problems.

Introduction to Quadratic Equation

The length of a rectangle is 3 cm more than its width. Its area is equal to 54 square centimeters. What is its length and width?

Solution

Let

x = width of rectangle
x + 3 = length of rectangle

The area of a rectangle is the product of the length and width, so we have

Area= x(x + 3)

which is equal to 54.

Therefore, we can form the following equation:

x(x + 3) = 54.

By the distributive property, we have

x^2 + 3x = 54

Finding the value of x

In the equation, we want to find the value of x that makes the equation true. Without algebraic manipulation, we can find the value of x by assigning various values to x. The equation x^2 + 3x = 54 indicates that one number is greater than the other by 3 and their product is 54. Examining the numbers with product as 54, we have,

1 and 54
2 and 27
3 and 18
6 and 9.

Note: We have excluded the negative (e.g. (-1)(-54) = 54) numbers since a side length cannot be negative.

Now, 9-6 = 3 which means that the side lengths of the rectangle are 6 and 9. Yes, their product is 54 and one is 3 greater than the other.

In the equation above, subtracting both sides by 54, we have

x^2 + 3x - 54 = 54 - 54

x^2 + 3x - 54 = 0.

The equation that we formed above is an example of a quadratic equation.

A quadratic equation is of the form ax^2 + bx + c = 0, where a, b, and c are real numbers and a not equal to 0. In the example above, a = 1, b = 3, and c = -54.

In the problem above, we got the value of x by testing several values, however, there are more systematic methods. In the next post, we will be discussing one of these methods. These methods are factoring, completing the square, and quadratic formula.

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