Chapter 5.12: Solving Quadratic Equations Using Substitution

Mike LePine

Chapter 5.12: 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 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 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.

Questions

Factor each of the following polynomials and solve what you can.

$x^4-5x^2+4=0$
$y^4-9y^2+20=0$
$m^4-7m^2-8=0$
$y^4-29y^2+100=0$
$a^4-50a^2+49=0$
$b^4-10b^2+9=0$
$x^4+64=20x^2$
$6z^6-z^3=12$
$z^6-216=19z^3$
$x^6-35x^3+216=0$

Answers to odd questions

1. $\text{let }u=x^2$
$\therefore u^2-5u+4=0$
$\text{factors to }(u-4)(u-1)=0$
$\text{replace }u: (x^2-4)(x^2-1)=0 \\$
$(x-2)(x+2)(x-1)(x+1)=0$
$x=\pm 2, \pm 1$

3. $u=m^2$
$\therefore u^2-7u-8=0$
$(u-8)(u+1)=0$
$(m^2-8)(m^2+1)=0 \\$
$(m+\sqrt{8})(m-\sqrt{8})(m^2+1)=0$
$m=\pm \sqrt{8}\text{ or }\pm 2\sqrt{2}$
$m^2+1\text{ has 2 non-real solutions}$

5. $\text{let }u=a^2$
$\therefore u^2-50u+49=0$
$(u-49)(u-1)=0$
$(a^2-49)(a^2-1)=0 \\$
$(a-7)(a+7)(a-1)(a+1)=0$
$a=\pm 7, \pm1$

7. $x^4-20x^2+64=0$
$\text{let }u=x^2$
$\therefore u^2-20u+64=0$
$(u-16)(u-4)=0$
$(x^2-16)(x^2-4)=0 \\$
$(x-4)(x+4)(x-2)(x+2)=0$
$x=\pm 4, \pm 2$

9. $z^6-19z^3-216=0$
$\text{let }u=z^3$
$\therefore u^2-19u-216=0$
$(u-27)(u+8)=0$
$(z^3-27)(z^3+8)=0 \\$
$(z-3)(z^2+3z+9)(z+2)(z^2-2z+4)=0$
$z=3, -2$
$2\text{ non-real solutions each for the 2nd and 4th factors}$

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