Learning Outcomes
- Solve equations reducible to quadratic form including using [latex] u [/latex]-substitution.
In this section we will learn to factor expressions which may not appear factorable at first, but after making a substitution they become factorable using our standard techniques.
As a warm-up, consider the following expression to factor:
[latex]x^4-25[/latex]
Both terms are squares since [latex] x^4 = (x^2)^2 [/latex], so we may use the difference of squares formula to factor this. However, we will first make a substitution to make it more clear that the formula applies. The substitution suggested by our above observation is [latex] u=x^2 [/latex]. After making this substitution, our expression becomes
[latex]\begin{align}x^4-25&=(x^2)^2-25\\ &=u^2-25\\ &=(u+5)(u-5)\\ &=(x^2+5)(x^2-5) \end{align}[/latex]
Make sure to write your final answer using the same variable that the problem started with.
Let’s use this technique now to solve an equation.
Example
Solve [latex]x^4-3x^2-4=0[/latex].
Show Solution
As in the previous example, we note that after a substitution of [latex] u=x^2 [/latex], the equation becomes quadratic. We can solve the resulting quadratic equation by factoring (or quadratic formula if necessary).
[latex]\begin{align}x^4-3x^2-4&=0\\ (x^2)^2-3(x^2)-4&=0\\ u^2-3u-4&=0\\ (u-4)(u+1)&=0\\ u-4=0 \textsf{ or } u+1=0 &\\ u=4 \textsf{ or } u=1& \end{align}[/latex]
Now substitute [latex] x^2 [/latex] for [latex] u [/latex] to solve for the original variable in each equation.
[latex]\begin{align}x^2&=4\\ x&=\pm \sqrt{4}\\ x&=2, -2 \end{align}[/latex]
[latex]\begin{align}x^2&=-1\\ x&=\pm \sqrt{-1}\\ x&=i, -i \end{align}[/latex]
The four complex solutions are [latex] x=2, -2, i, -i [/latex].
Since the equation was possible to change into quadratic by an appropriate substitution, we say that the original equation was quadratic in form.
In our remaining examples we will show more examples of substitutions that result in a quadratic equation to solve. Observe that it can be helpful to write the starting expression in descending order (in some sense of the word “descending”) and using the middle term to determine the correct substitution.
Example
Solve [latex]x^{-2}+5x^{-1}+6=0[/latex].
Show Solution
If the exponents on this trinomial were positive, we could factor this as [latex](x+2)(x+3)[/latex].
This suggests using a substitution [latex] u=x^{-1} [/latex]. We verify that that [latex] u^2 = (x^{-1})^2 = x^{-2} [/latex], which allows us to substitute the first term as well. Now we obtain
[latex]\begin{align}x^{-2}+5x^{-1}+6&=0\\ (x^{-1})^2+5(x^{-1})+6&=0\\ u^2+5u+6&=0\\ (u+2)(u+3)&=0\\ u+2=0 \textsf{ or } u+3=0 &\\ u=-2 \textsf{ or } u=-3& \end{align}[/latex]
Now substitute [latex] x^{-1} [/latex] for [latex] u [/latex] to solve for the original variable in each equation.
[latex]\begin{align}x^{-1}&=-2\\ \dfrac{1}{x}&=-2\\ 1&=-2x\\ x&=-\dfrac{1}{2} \end{align}[/latex]
[latex]\begin{align}x^{-1}&=-3\\ \dfrac{1}{x}&=-3\\ 1&=-3x\\ x&=-\dfrac{1}{3} \end{align}[/latex]
The two solutions are [latex] x=-\dfrac{1}{2} [/latex] or [latex] x=-\dfrac{1}{3} [/latex].
Note that sometimes there are alternative approaches to solving these problems. We could have written our starting problem as [latex] \dfrac{1}{x^2}+\dfrac{5}{x}+6=0 [/latex] and solved using our rational equation methods of the last chapter, multiplying both sides by the LCM of [latex] x^2 [/latex] to obtain [latex] 1+5x+6x^2=0 [/latex]. You can verify that this results in the same solutions found above.
Example
Solve [latex]\sqrt{x}-x=-2[/latex].
Show Solution
We begin by getting everything on the left side. Since a square root is equivalent to an exponent of [latex] \dfrac{1}{2} [/latex], this suggests a “descending order” to place the terms in with the square root as the middle term.
[latex]\begin{align}\sqrt{x}-x&=-2\\ \sqrt{x}-x+2&=0\\ -x+\sqrt{x}+2&=0 \end{align}[/latex]
The middle term suggests using the substitution [latex] u=\sqrt{x} [/latex]. We verify that [latex] u^2=(\sqrt{x})^2=x [/latex], which replaces the first term. Thus we continue
[latex]\begin{align} -u^2+u+2&=0\\ -(u^2-u-2)&=0\\ -(u-2)(u+1)&=0\\ u-2=0 \textsf{ or } u+1=0 &\\ u=2 \textsf{ or } u=-1& \end{align}[/latex]
Now substitute [latex] \sqrt{x} [/latex] for [latex] u [/latex] to solve for the original variable in each equation.
[latex]\begin{align}\sqrt{x}&=2\\ (\sqrt{x})^2&=2^2\\ x&=4 \end{align}[/latex]
[latex]\sqrt{x}=-1[/latex]
The second equation cannot be solved, as the principal square root is always positive. Thus the solution is [latex] x=4 [/latex].
Summary
In this section, we used a substitution to turn a variety of different equations into quadratic equations which could be solved either by factoring or using the previous methods of this chapter.
