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8. Solving Quadratic Equations
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United States of AmericaFL
Grade 912

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

MA.912.AR.1.1

Identify and interpret parts of an equation or expression that represent a quantity in terms of a mathematical or real-world context, including viewing one or more of its parts as a single entity.

MA.912.AR.1.7

Rewrite a polynomial expression as a product of polynomials over the real number system.

MA.912.AR.2.1

Given a real-world context, write and solve one-variable multi-step linear equations.

MA.912.AR.3.1

Given a mathematical or real-world context, write and solve one-variable quadratic equations over the real number system.

MA.912.AR.3.8

Solve and graph mathematical and real-world problems that are modeled with quadratic functions. Interpret key features and determine constraints in terms of the context.

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