10.5 Solving Quadratic Equations Using Substitution

Terrance Berg

Chapter 10: Quadratics

10.5 Solving Quadratic Equations Using Substitution

Factoring trinomials in which the leading term is not 1 is only slightly more difficult than when the leading coefficient is 1. The method used to factor the trinomial is unchanged.

Example 10.5.1

Solve for $x$ in $x^4 - 13x^2 + 36 = 0$ .

First start by converting this trinomial into a form that is more common. Here, it would be a lot easier when factoring $x^2 - 13x + 36 = 0.$ There is a standard strategy to achieve this through substitution.

First, let $u = x^2$ . Now substitute $u$ for every $x^2$ , the equation is transformed into $u^2-13u+36=0$ .

$u^2 - 13u + 36 = 0$ factors into $(u - 9)(u - 4) = 0$ .

Once the equation is factored, replace the substitutions with the original variables, which means that, since $u = x^2$ , then $(u - 9)(u - 4) = 0$ becomes $(x^2 - 9)(x^2 - 4) = 0$ .

To complete the factorization and find the solutions for $x$ , then $(x^2 - 9)(x^2 - 4) = 0$ must be factored once more. This is done using the difference of squares equation: $a^2 - b^2 = (a + b)(a - b)$ .

Factoring $(x^2 - 9)(x^2 - 4) = 0$ thus leaves $(x - 3)(x + 3)(x - 2)(x + 2) = 0$ .

Solving each of these terms yields the solutions $x = \pm 3, \pm 2$ .

This same strategy can be followed to solve similar large-powered trinomials and binomials.

Example 10.5.2

Factor the binomial $x^6 - 7x^3 - 8 = 0$ .

Here, it would be a lot easier if the expression for factoring was $x^2 - 7x - 8 = 0$ .

First, let $u = x^3$ , which leaves the factor of $u^2 - 7u - 8 = 0$ .

$u^2 - 7u - 8 = 0$ easily factors out to $(u - 8)(u + 1) = 0$ .

Now that the substituted values are factored out, replace the $u$ with the original $x^3$ . This turns $(u - 8)(u + 1) = 0$ into $(x^3 - 8)(x^3 + 1) = 0$ .

The factored $(x^3 - 8)$ and $(x^3 + 1)$ terms can be recognized as the difference of cubes.

These are factored using $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ and $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ .

And so, $(x^3 - 8)$ factors out to $(x - 2)(x^2 + 2x + 4)$ and $(x^3 + 1)$ factors out to $(x + 1)(x^2 - x + 1)$ .

Combining all of these terms yields:

$(x - 2)(x^2 + 2x + 4)(x + 1)(x^2 - x + 1) = 0$

The two real solutions are $x = 2$ and $x = -1$ . Checking for any others by using the discriminant reveals that all other solutions are complex or imaginary solutions.