In Section solveeqnsbyfactoring, we reviewed how to solve basic non-linear equations by factoring. The astute reader should have noticed that all of the equations in that section were carefully constructed so that the polynomials could be factored using the integers. To demonstrate just how contrived the equations had to be, we can solve \(2x^2+5x-3=0\) by factoring, \((2x-1)(x+3) = 0\), from which we obtain \(x = \frac {1}{2}\) and \(x = -3\). If we change the \(5\) to a \(6\) and try to solve \(2x^2 + 6x - 3 = 0\), however, we find that this polynomial doesn’t factor over the integers and we are stuck. It turns out that there are two real number solutions to this equation, but they are irrational numbers, and the goal of this section is to review the techniques which allow us to find these solutions. In this section, we focus our attention on quadratic equations.

Think of quadratic equations as equations that are one degree up from linear equations - instead of the highest power of \(x\) being just \(x = x^1\), it’s \(x^2\). The simplest class of quadratic equations to solve are the ones in which \(b = 0\). In that case, we have the following.

Solving Quadratic Equations by Extracting Square Roots

If \(c\) is a real number with \(c \geq 0\), the solutions to \(x^2 = c\) are \(x = \pm \sqrt {c}\).

Note: If \(c < 0\), \(x^2 = c\) has no real number solutions.

There are a couple different ways to see why Extracting Square Roots works, both of which are demonstrated by solving the equation \(x^2 = 3\). If we follow the procedure outlined in the previous section, we subtract \(3\) from both sides to get \(x^2 - 3 = 0\) and we now try to factor \(x^2 - 3\). As mentioned in the remarks following Definition irreduciblepoly, we could think of \(x^2 - 3 = x^2 - (\sqrt {3})^2\) and apply the Difference of Squares formula to factor \(x^2 - 3 = (x-\sqrt {3})(x+\sqrt {3})\). We solve \((x-\sqrt {3})(x+\sqrt {3}) = 0\) by using the Zero Product Property as before by setting each factor equal to zero: \(x - \sqrt {3} = 0\) and \(x+\sqrt {3} - 0\). We get the answers \(x = \pm \sqrt {3}\). In general, if \(c \geq 0\), then \(\sqrt {c}\) is a real number, so \(x^2 - c = x^2 - (\sqrt {c})^2 = (x-\sqrt {c})(x+\sqrt {c})\). Replacing the ‘\(3\)’ with ‘\(c\)’ in the above discussion gives the general result.

Another way to view this result is to visualize ‘taking the square root’ of both sides: since \(x^2 = c\), \(\sqrt {x^2} = \sqrt {c}\). How do we simplify \(\sqrt {x^2}\)? We have to exercise a bit of caution here. Note that \(\sqrt {(5)^2}\) and \(\sqrt {(-5)^2}\) both simplify to \(\sqrt {25} = 5\). In both cases, \(\sqrt {x^2}\) returned a positive number, since the negative in \(-5\) was ‘squared away’ before we took the square root. In other words, \(\sqrt {x^2}\) is \(x\) if \(x\) is positive, or, if \(x\) is negative, we make \(x\) positive - that is, \(\sqrt {x^2} = |x|\), the absolute value of \(x\). So from \(x^2 = 3\), we ‘take the square root’ of both sides of the equation to get \(\sqrt {x^2} = \sqrt {3}\). This simplifies to \(|x| = \sqrt {3}\), which by Theorem absvalequality is equivalent to \(x = \sqrt {3}\) or \(x = -\sqrt {3}\). Replacing the ‘\(3\)’ in the previous argument with ‘\(c\),’ gives the general result.

As you might expect, Extracting Square Roots can be applied to more complicated equations. Consider the equation below. We can solve it by Extracting Square Roots provided we first isolate the quantity that is being squared :

\[ \begin{array}{rclr} 2\left (x + \frac {3}{2}\right )^2 - \frac {15}{2} & = & 0 & \\ 2\left (x + \frac {3}{2}\right )^2 & = & \frac {15}{2} & \text {Add $\frac {15}{2}$} \\ \left (x + \frac {3}{2}\right )^2 & = & \frac {15}{4} & \text {Divide by $2$} \\ x + \frac {3}{2} & = & \pm \sqrt {\frac {15}{4}} & \text {Extract Square Roots} \\ x + \frac {3}{2} & = & \pm \frac {\sqrt {15}}{2} & \text {Property of Radicals} \\ x & = & -\frac {3}{2} \pm \frac {\sqrt {15}}{2} & \text {Subtract $\frac {3}{2}$} \\ x & = & -\frac {3 \pm \sqrt {15}}{2} & \text {Add fractions} \\ \end{array} \]

Let’s return to the equation \(2x^2 + 6x - 3 = 0\) from the beginning of the section. We leave it to the reader to expand the left side and show that

\[2\left (x + \frac {3}{2}\right )^2 - \frac {15}{2} = 2x^2 + 6x - 3. \]

In other words, we can solve \(2x^2 + 6x - 3 = 0\) by transforming into an equivalent equation. This process, you may recall, is called ‘Completing the Square.’ We’ll revisit Completing the Square in Section QuadraticFunctions in more generality and for a different purpose but for now we revisit the steps needed to complete the square to solve a quadratic equation.

Solving Quadratic Equations: Completing the Square

To solve a quadratic equation \(ax^2 + bx + c = 0\) by Completing the Square:

  1. Subtract the constant \(c\) from both sides.
  2. Divide both sides by \(a\), the coefficient of \(x^2\). (Remember: \(a \neq 0\).)
  3. Add \(\left (\frac {b}{2a}\right )^2\) to both sides of the equation. (That’s half the coefficient of \(x\), squared.)
  4. Factor the left hand side of the equation as \(\left (x + \frac {b}{2a}\right )^2\).
  5. Extract Square Roots.
  6. Subtract \(\frac {b}{2a}\) from both sides.

To refresh our memories, we apply this method to solve \(3x^2 - 24x + 5 = 0\):

\[ \begin{array}{rclr} 3x^2 - 24x + 5 & = & 0 & \\ 3x^2 - 24x & = & -5 & \text {Subtract $c = 55$} \\ x^2 - 8x & = & -\frac {5}{3} & \text {Divide by $a = 3$} \\ x^2 - 8x + 16 & = & -\frac {5}{3} + 16 & \text {Add $\left (\frac {b}{2a}\right )^2 = (-4)^2 = 16$} \\ (x - 4)^2 & = & \frac {43}{3} & \text {Factor: Perfect Square Trinomial} \\ x - 4 & = & \pm \sqrt {\frac {43}{3}} & \text {Extract Square Roots} \\ x & = & 4 \pm \sqrt {\frac {43}{3}} & \text {Add $4$} \\ \end{array}\]

At this point, we use properties of fractions and radicals to ‘rationalize’ the denominator:

\[ \sqrt {\frac {43}{3}} = \sqrt {\frac {43 \cdot 3}{3 \cdot 3}} = \frac {\sqrt {129}}{\sqrt {9}} = \frac {\sqrt {129}}{3} \]

We can now get a common (integer) denominator which yields:

\[x= 4 \pm \sqrt {\frac {43}{3}} = 4 \pm \frac {\sqrt {129}}{3} = \frac {12 \pm \sqrt {129}}{3} \]

The key to Completing the Square is that the procedure always produces a perfect square trinomial. To see why this works every single time, we start with \(ax^2 + bx + c = 0\) and follow the procedure:

\[ \begin{array}{rclr} ax^2 + bx + c & = & 0 & \\ ax^2 + bx & = & -c & \text {Subtract $c$} \\ x^2 + \frac {bx}{a} & = & -\frac {c}{a} & \text {Divide by $a \neq 0$} \\ x^2 + \frac {bx}{a} + \left (\frac {b}{2a}\right )^2 & = & -\frac {c}{a} + \left (\frac {b}{2a}\right )^2 & \text {Add $ \left (\frac {b}{2a}\right )^2$} \\ \end{array} \]

(Hold onto the line above for a moment.) Here’s the heart of the method - we need to show that

\[ x^2 + \frac {bx}{a} + \left (\frac {b}{2a}\right )^2 = \left (x + \frac {b}{2a}\right )^2 \]

To show this, we start with the right side of the equation and apply the Perfect Square Formula from Theorem SpecialProducts

\[ \left (x + \frac {b}{2a}\right )^2 = x^2 + 2\left (\frac {b}{2a}\right )x + \left (\frac {b}{2a}\right )^2 = x^2 + \frac {bx}{a} + \left (\frac {b}{2a}\right )^2 \, \checkmark \]

With just a few more steps we can solve the general equation \(ax^{2} + bx + c = 0\) so let’s pick up the story where we left off. (The line on the previous page we told you to hold on to.)

\[ \begin{array}{rclr} x^2 + \frac {bx}{a} + \left (\frac {b}{2a}\right )^2 & = & -\frac {c}{a} + \left (\frac {b}{2a}\right )^2 & \\ \left (x + \frac {b}{2a}\right )^2 & = & -\frac {c}{a} + \frac {b^2}{4a^2} & \text {Factor: Perfect Square Trinomial} \\ \left (x + \frac {b}{2a}\right )^2 & = & -\frac {4ac}{4a^2} + \frac {b^2}{4a^2} & \text {Get a common denominator}\\ \left (x + \frac {b}{2a}\right )^2 & = & \frac {b^2 - 4ac}{4a^2} & \text {Add fractions}\\ x + \frac {b}{2a} & = & \pm \sqrt {\frac {b^2 - 4ac}{4a^2}} & \text {Extract Square Roots} \\ x + \frac {b}{2a} & = & \pm \frac {\sqrt {b^2 - 4ac}}{2a} & \text {Properties of Radicals} \\ x & = & - \frac {b}{2a} \pm \frac {\sqrt {b^2 - 4ac}}{2a} & \text {Subtract $\frac {b}{2a}$} \\ x & = & \frac {-b \pm \sqrt {b^2 - 4ac}}{2a} & \text {Add fractions.} \\ \end{array}\]

Lo and behold, we have derived the legendary Quadratic Formula!

We can check our earlier solutions to \(2x^2 + 6x - 3 = 0\) and \(3x^2 - 24x + 5 = 0\) using the Quadratic Formula. For \(2x^2 + 6x - 3 = 0\), we identify \(a = 2\), \(b = 6\) and \(c = -3\). The quadratic formula gives:

\[ x = \frac {-6 \pm \sqrt {6^2 - 4(2)(-3)}}{2(2)} - \frac {-6 \pm \sqrt {36 + 24}}{4} = \frac {-6 \pm \sqrt {60}}{4} \]

Using properties of radicals (\(\sqrt {60} = 2 \sqrt {15}\)), this reduces to \(\frac {2(-3 \pm \sqrt {15})}{4} =\frac {-3 \pm \sqrt {15}}{2}\). We leave it to the reader to show these two answers are the same as \(-\frac {3 \pm \sqrt {15}}{2}\), as required.

For \(3x^2 - 24x + 5 = 0\), we identify \(a = 3\), \(b = -24\) and \(c = 5\). Here, we get:

\[ x = \frac {-(-24) \pm \sqrt {(-24)^2 - 4(3)(5)}}{2(3)} = \frac {24 \pm \sqrt {516}}{6} \]

Since \(\sqrt {516} = 2\sqrt {129}\), this reduces to \(x = \frac {12 \pm \sqrt {129}}{3}\).

It is worth noting that the Quadratic Formula applies to all quadratic equations - even ones we could solve using other techniques. For example, to solve \(2x^2 + 5x - 3 = 0\) we identify \(a = 2\), \(b = 5\) and \(c = -3\). Plugging those into the Quadratic Formula yields:

\[ x = \frac {-5 \pm \sqrt {5^2 - 4(2)(-3)}}{2(2)} = \frac {-5 \pm \sqrt {49}}{4} = \frac {-5 \pm 7}{4} \]

At this point, we have \(x = \frac {-5+7}{4} = \frac {1}{2}\) and \(x = \frac {-5-7}{4} = \frac {-12}{4} = -3\) - the same two answers we obtained factoring. We can also use it to solve \(x^2 = 3\), if we wanted to. From \(x^2 -3 = 0\), we have \(a = 1\), \(b = 0\) and \(c = -3\). The Quadratic Formula produces

\[ x = \frac {-0 \pm \sqrt {0^2 - 4(1)(3)}}{2(1)} = \frac {\pm \sqrt {12}}{2} = \pm \frac {2\sqrt {3}}{2} = \pm \sqrt {3}\]

As this last example illustrates, while the Quadratic Formula can be used to solve every quadratic equation, that doesn’t mean it should be used. Many times other methods are more efficient. We now provide a more comprehensive approach to solving Quadratic Equations.

Strategies for Solving Quadratic Equations
  • If the variable appears in the squared term only, isolate it and Extract Square Roots.

  • Otherwise, put the nonzero terms on one side of the equation so that the other side is \(0\).

    • Try factoring.
    • If the expression doesn’t factor easily, use the Quadratic Formula.

The reader is encouraged to pause for a moment to think about why ‘Completing the Square’ doesn’t appear in our list of strategies despite the fact that we’ve spent the majority of the section so far talking about it. Let’s get some practice solving quadratic equations, shall we?

Our last example above, the ‘Quadratic in Disguise’, hints that the Quadratic Formula is applicable to a wider class of equations than those which are strictly quadratic. We give some general guidelines to recognizing these beasts in the wild on the next page.

Identifying Quadratics in Disguise

An equation is a ‘Quadratic in Disguise’ if it can be written in the form: \(ax^{2m} + bx^{m} + c = 0\).

In other words:

  • There are exactly three terms, two with variables and one constant term.
  • The exponent on the variable in one term is exactly twice the variable on the other term.

To transform a Quadratic in Disguise to a quadratic equation, let \(u = x^m\) so \(u^2 = (x^m)^2 = x^{2m}\). This transforms the equation into \(au^2 + bu + c = 0\).

For example, \(3x^6 - 2x^3 + 1 = 0\) is a Quadratic in Disguise, since \(6 = 2 \cdot 3\). If we let \(u = x^3\), we get \(u^2 = (x^3)^2 = x^6\), so the equation becomes \(3u^2 - 2u + 1 = 0\). However, \(3x^6 - 2x^2 + 1 = 0\) is not a Quadratic in Disguise, since \(6 \neq 2\cdot 2\). The substitution \(u = x^2\) yields \(u^2 = (x^2)^2 = x^4\), not \(x^6\) as required. We’ll see more instances of ‘Quadratics in Disguise’ in later sections.

We close this section with a review of the discriminant of a quadratic equation as defined below.

The discriminant is the radicand of the square root in the quadratic formula:

\[x = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a} \]

It discriminates between the nature and number of solutions we get from a quadratic equation. The results are summarized below.

For example, the equation \(x^2 + x - 1 = 0\) has two real number solutions since the discriminant works out to be \((1)^2 - 4(1)(-1) = 5 > 0\). This results in a \(\pm \sqrt {5}\) in the Quadratic Formula which then generates two different answers. On the other hand, \(x^2 + x + 1 = 0\) has no real solutions since here, the discriminant is \((1)^2 - 4(1)(1) = -3 < 0\) which generates a \(\pm \sqrt {-3}\) in the Quadratic Formula. The equation \(x^2 + 2x +1 = 0\) has discriminant \((2)^2 - 4(1)(1) = 0\) so in the Quadratic Formula we get a \(\pm \sqrt {0} = 0\) thereby generating just one solution. More can be said as well. For example, the discriminant of \(6x^2 - x - 40 = 0\) is \(961\). This is a perfect square, \(\sqrt {961} = 31\), which means our solutions are rational numbers. When our solutions are rational numbers, the quadratic actually factors nicely. In our example \(6x^2 - x - 40 = (2x+5)(3x-8)\). Admittedly, if you’ve already computed the discriminant, you’re most of the way done with the problem and probably wouldn’t take the time to experiment with factoring the quadratic at this point – but we’ll see another use for this analysis of the discriminant in Example rationalexpressionreviewex.