Topics
11. Quadratic Equations
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United States of America
Grades 9-12

11.05 Solving using appropriate methods

Lesson
Practice

Introduction

We explored several methods for solving quadratic equations in the past four lessons. However, it may be difficult to know when it is best, or easiest, to use each method. In this lesson, we will discuss the advantages and disadvantages of each method to help us determine when it is easiest to use each.

Solving quadratic equations using appropriate methods

We have several methods we can use to solve quadratic equations. To determine which method is the most suitable we need to look at the form of the quadratic equation. Some are easily solved by factoring for example, or by taking the square root.

If we are unable to solve the quadratic easily using one of these methods, the quadratic formula is often the best approach since it can be used to solve any quadratic equation once it's written in standard form. If we have access to technology, drawing the graph of the corresponding quadratic function can help us find exact solutions or approximate a solution if it is not an integer value.

Advantages: Helps us visualize the quadratic and its key features

Disadvantages: Only best when intercepts are integers, in which case it could have been factored instead

Equation form: Any form is fine if using techonology, otherwise it is best in a form that is equal to 0

Advantages: This is usually the fastest method

Disadvantages: Not all polynomials are factorable, some factorable polynomials are difficult to factor

Equation form:

ax^2+bx+c=0 where a,b,c are small

Advantages: Simplest method for solving equations in vertex form or equations missing an x-term

Disadvantages: Few equations are given in this form

Equation form: x^2=k or a(x-h)^2=k

Advantages: Can be used to solve any quadratic equation

Disadvantages: Requires more steps than other methods, fractions make it difficult

Equation form: x^2+bx+c=0 where b is even

Advantages: Can be used to solve any quadratic equation

Disadvantages: Can be time-consuming, many opportunities to make miscalculations, although calculators simplify its use

Equation form:

ax^2+bx+c=0 where a,b,c are large

There is not one correct method for solving a quadratic equation. You would not be wrong by using one method over another; it is just easier, sometimes more practical, to use some methods over others.

Examples

Example 1

For the following quadratic equations, find the solution using an efficient method. Justify which method you used.

a

x^2-7x+12=0

Worked Solution
Create a strategy

The leading coefficient of x^2 is 1, so we can check if this can be easily factored. The prime factors of 12 are \pm 1,\, \pm 2,\, \pm3,\,\pm4,\, \pm6,\, \pm 12, and we want to find two factors that have a product of 12 and sum to -7. As the product is positive but the sum is negative, we know both factors must be negative.

Apply the idea

Since the equation can be factored by grouping, we will factor the equation and solve it.

The two factors that have a product of 12 and a sum of -7 are -3 and -4. We can write the equation in factored form as \left(x-4\right)\left(x-3\right)=0, which gives us two solutions x=3 and x=4.

Reflect and check

In general, if the coefficients are small, and especially if a=1, it is worth checking to see if we can easily factor the equation to solve.

b

x^2-11=21

Worked Solution
Create a strategy

Here we have b=0, and can easily isolate the x^2, which means we can solve this by using square roots.

Apply the idea

Since we can easily isolate x^2, we will solve the equation using square roots as follows:

\displaystyle x^2-11\displaystyle =\displaystyle 21Given equation
\displaystyle x^2\displaystyle =\displaystyle 32Add 11 to both sides
\displaystyle x\displaystyle =\displaystyle \pm \sqrt{32}Evaluate the square root of both sides
\displaystyle x\displaystyle =\displaystyle \pm \sqrt{16\cdot2}Factor 32
\displaystyle x\displaystyle =\displaystyle \pm \sqrt{16}\sqrt{2}Multiplication property of radicals
\displaystyle x\displaystyle =\displaystyle \pm 4\sqrt{2}Evaluate the radical

x=\pm 4\sqrt{2}, giving us two solutions: x=4\sqrt{2} and x=-4\sqrt{2}

Reflect and check

In general, if we can easily rearrange the equation into the form \left(x-h\right)^2=k for some positive value of k then solving using square roots is a suitable method.

c

3x^2-24x+20=5

Worked Solution
Create a strategy

For most of the methods we know, the quadratic needs to be equal to zero first. We can subtract 5 from both sides, then check see if factoring can be used.

Apply the idea

Since the trinomial is equal to a constant, we will first set the equation equal to zero and attempt to factor the trinomial. Then, we can determine an approach that is appropriate for solving this equation.

\displaystyle 3x^2-24x+20\displaystyle =\displaystyle 5Given equation
\displaystyle 3x^2-24x+15\displaystyle =\displaystyle 0Subtract 5 from both sides
\displaystyle 3(x^2-8x+5)\displaystyle =\displaystyle 0Factor the GCF of 3
\displaystyle x^2-8x+5\displaystyle =\displaystyle 0Divide by 3 on both sides

From here, we can see that the equation cannot be factored further. Since a=1 and b is even, we can use completing the square to solve.

\displaystyle x^2-8x\displaystyle =\displaystyle -5Subtraction property of equality
\displaystyle x^2-8x+16\displaystyle =\displaystyle -5+16Complete the square
\displaystyle (x-4)^2\displaystyle =\displaystyle 11Factor the left side, evaluate the right side
\displaystyle x-4\displaystyle =\displaystyle \pm \sqrt{11}Evaluate the square root of both sides
\displaystyle x\displaystyle =\displaystyle 4\pm\sqrt{11}Add 4 to both sides
Reflect and check

The quadratic formula could have been used, but it may have been more time-consuming, especially if we didn’t factor out the GCF first.

\displaystyle 3x^2-24x+15\displaystyle =\displaystyle 0Original equation set equal to 0
\displaystyle x\displaystyle =\displaystyle \dfrac{-(-24)\pm\sqrt{(-24)^2-4(3)(15)}}{2(3)}Substitute a,b,c into quadratic formula
\displaystyle =\displaystyle \dfrac{24\pm\sqrt{396}}{6}Evaluate the operations
\displaystyle =\displaystyle \dfrac{24\pm\sqrt{36}\sqrt{11}}{6}Multiplication property of radicals
\displaystyle =\displaystyle \dfrac{24\pm 6\sqrt{11}}{6}Evaluate the radical
\displaystyle =\displaystyle 4\pm\sqrt{11}Evaluate the division

Example 2

A rectangular enclosure is to be constructed from 100 meters of wooden fencing. The area of the enclosure is given by A = 50 x - x^{2}, where x is the length of one side of the rectangle. If the area is 525 \text{ m}^2, determine the side lengths.

Worked Solution
Create a strategy

We can set up and solve a quadratic equation, 50x-x^{2}=525. Since the values are large we will try solving this problem with the quadratic formula. The two solutions will be the side lengths of the enclosure.

Apply the idea

Rearranging the equation into standard form we get x^2-50x+525=0.

We can solve this using the quadratic equation:

\displaystyle x\displaystyle =\displaystyle \dfrac{-b \pm \sqrt{b^2-4ac}}{2a}Quadratic formula
\displaystyle x\displaystyle =\displaystyle \dfrac{-\left(-50\right) \pm \sqrt{\left(-50\right)^2-4\left(1\right)\left(525\right)}}{2\left(1\right)}Substitute a=1,b=-50,c=525
\displaystyle x\displaystyle =\displaystyle \dfrac{50 \pm \sqrt{400}}{2}Evaluate the operations
\displaystyle x\displaystyle =\displaystyle \dfrac{50 \pm 20}{2}Evaluate the square root

This leaves us with two values, x=\dfrac{50+20}{2} and x=\dfrac{50-20}{2}. Evaluating each expression for x we get x=35 and x=15 as the side lengths of the rectangular enclosure.

Reflect and check

We can confirm our answer is correct by checking the conditions of the problem. We had 100 meters of fencing and 2\left(35+15\right)=100. We needed the area to be 525 \text{ m}^2 and 35\left(15\right)=525 as required.

Since there are two rational solutions, the quadratic equation was also factorable:

x^2-50x+525=\left(x-35\right)\left(x-15\right), but these factors are not immediately obvious.

Idea summary

Below is a list of the easiest method to use and the form of the quadratic equation for which we should use it:

Easiest equation form:
Graphing\text{Any form is fine when using technology}
Factoringax^2+bx+c=0\text{ where }a,b,c\text{ are small}
Square root propertyx^2=k\text{ or }a(x-h)^2=k
Completing the squarex^2+bx+c=0\text{ where }b\text{ is even}
Quadratic formulaax^2+bx+c=0\text{ where }a,b,c\text{ are large }

Outcomes

A.CED.A.1

Create equations and inequalities in one variable and use them to solve problems.

A.CED.A.2

Create equations in two or more variables to represent relationships between quantities; graph equations on coordinate axes with labels and scales.

A.REI.B.4

Solve quadratic equations in one variable.

A.REI.B.4.A

Use the method of completing the square to transform any quadratic equation in x into an equation of the form (x - p)^2 = q that has the same solutions. Derive the quadratic formula from this form.

A.REI.B.4.B

Solve quadratic equations by inspection (e.g. For x^2 = 49), taking square roots, completing the square, the quadratic formula and factoring, as appropriate to the initial form of the equation. Recognize when the quadratic formula gives complex solutions and write them as a ± bi for real numbers a and b.

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