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8. Solving Quadratic Equations
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United States of AmericaTennessee
Algebra 1

8.02 Solving quadratic equations by factoring

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Concept summary

The zero product property states that if a product of two or more factors is equal to 0, then at least one of the factors must be equal to 0. That is, if we know that xy=0 then at least one of x=0 or \\y=0 must be true.

We can use this property to solve quadratic equations by first writing the equation in factored form: a\left(x-x_1\right)\left(x-x_2\right)=0

If we can write a quadratic equation in the factored form, then we know that either x-x_1=0 or \\ x-x_2=0. This means that the solutions to the quadratic equation are x_1 and x_2. This approach can be useful if the equation has rational solutions.

If an equation is in the form \left(x-x_1\right)^2=0 then it has only one real solution, x=x_1.

We can use the following process for solving most quadratic equations:

  1. Gather together all the terms on one side of the equation so that the other side is equal to 0
  2. Factor these terms using any appropriate techniques.
  3. Use the zero product property to split the quadratic equation into two linear equations.
  4. Solve the linear equations to get the solutions to the original quadratic equation

Worked examples

Example 1

Solve x^2+6x-55=0 by factoring:

Approach

First we want to find the factors of the constant term - 55 then choose the pair of factors (let's call them p and q) that add up to 6. After finding p and q, we factor the quadratic by writing it in the form \left(x + p\right)\left(x + q\right)=0.

The prime factors of -55 are \pm1,\, \pm 5,\, \pm11 and \pm 55, and we want a factor pair whose sum is 6.

Solution

The factor pair whose sum is 6 is -5 and 11, so we have p=-5, q=11.

Factoring gives us x^2+6x-55=\left(x-5\right)\left(x+11\right) which leads to the equation \left(x-5\right)\left(x+11\right)=0

We can then solve the equation by setting each factor equal to zero, giving us x-5=0 and x+11=0, which gives the solutions x=5 and x=-11.

Reflection

It is important to pay attention to the sign of the terms when determining the solution from a quadratic in factored form. In the example above the term of \left(x-5\right) leads to a solution of x=5 and \left(x+11\right) to a solution of x=-11.

It is also important to note that in the two forms \left(x+p\right)\left(x+q\right) and \left(x-x_1\right)\left(x-x_2\right), we can see that x_1=-p, and x_2=-q.

Example 2

The graph of a quadratic function has x-intercepts at \left(3,\,0\right) and \left(-1,\,0\right) and passes through the point \left(4,\,10\right). Write an equation in factored form that models this quadratic.

Approach

To write the equation for this quadratic in factored form we need to first identify the solutions of the equation. We can then substitute these values for x_1 and x_2. Next, we will need to find a by substituting one other point and solving the resulting equation.

The roots of the equation are at \left(-1,\,0\right) and \left(3,\,0\right) so we know the equation has solutions of -1 and 3.

Solution

Since the solutions are -1 and 3, we can write the factored form as:

y=a(x+1)(x-3)

for some value of a. We can find a by substituting in the coordinates of the additional point, \left(4,\,10\right), into the function.

To find a:

\displaystyle y\displaystyle =\displaystyle a(x+1)(x-3)Factored form
\displaystyle 10\displaystyle =\displaystyle a(4+1)(4-3)Substitute in the known values
\displaystyle 10\displaystyle =\displaystyle a(5)(1)Simplify the factors
\displaystyle 10\displaystyle =\displaystyle 5aSimplify
\displaystyle 2\displaystyle =\displaystyle aDivide both sides by 5

The equation of the quadratic function in factored form:

y=2(x+1)(x-3)

Outcomes

A1.A.CED.A.1

Create equations and inequalities in one variable and use them to solve problems in a real-world context.*

A1.A.CED.A.2

Create equations in two variables to represent relationships between quantities and use them to solve problems in a real-world context. Graph equations with two variables on coordinate axes with labels and scales, and use the graphs to make predictions.*

A1.A.CED.A.3

Create individual and systems of equations and/or inequalities to represent constraints in a contextual situation, and interpret solutions as viable or non-viable.*

A1.A.REI.A.1

Understand solving equations as a process of reasoning and explain the reasoning. Construct a viable argument to justify a solution method.

A1.A.REI.B.3

Solve quadratic equations and inequalities in one variable.

A1.A.REI.B.3.A

Solve quadratic equations by inspection (e.g., for x^2 = 49), taking square roots, knowing and applying the quadratic formula, and factoring, as appropriate to the initial form of the equation. Recognize when a quadratic equation has solutions that are not real numbers.

A1.MP1

Make sense of problems and persevere in solving them.

A1.MP3

Construct viable arguments and critique the reasoning of others.

A1.MP6

Attend to precision.

A1.MP7

Look for and make use of structure.

A1.MP8

Look for and express regularity in repeated reasoning.

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