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15.S: The Quadratic Formula

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    Previous Essential Skills sections introduced many of the mathematical operations you need to solve chemical problems. We now introduce the quadratic formula, a mathematical relationship involving sums of powers in a single variable that you will need to apply to solve some of the problems in this chapter.

    The Quadratic Formula

    Mathematical expressions that involve a sum of powers in one or more variables (e.g., \(x\)) multiplied by coefficients (such as \(a\)) are called polynomials. Polynomials of a single variable have the general form

    \[a_nx^n + ... + a_2x^2 + a_1x + a_0 \tag{15S.1}\]

    The highest power to which the variable in a polynomial is raised is called its order. Thus the polynomial shown here is of the nth order. For example, if \(n\) were 3, the polynomial would be third order.

    A quadratic equation is a second-order polynomial equation in a single variable x:

    \[ax^2 + bx + c = 0\tag{15S.2}\]

    According to the fundamental theorem of algebra, a second-order polynomial equation has two solutions—called roots—that can be found using a method called completing the square. In this method, we solve for \(x\) by first adding −c to both sides of the quadratic equation and then divide both sides by \(a\):

    \[x^2+bax =−ca\tag{15S.3}\]

    We can convert the left side of this equation to a perfect square by adding \(b^2/4a^2\), which is equal to \((b/2a)^2\):

    Left side: \[x^2+bax+b^24a^2=(x+b^2a)^2 \tag{15S.4}\]

    Having added a value to the left side, we must now add that same value, \(b^2 ⁄ 4a^2\), to the right side:

    \[(x+b^2a)^2=−ca+b24a^2 \tag{15S.5}\]

    The common denominator on the right side is \(4a^2\). Rearranging the right side, we obtain the following:

    \[(x+b^2a)^2=b^2−4ac4a^2 \tag{15S.6}\]

    Taking the square root of both sides and solving for x,

    \[x+b^2a= \dfrac{\pm \sqrt{b^2−4ac}}{2a} \tag{15S.7}\]

    \[x= \dfrac{−b \pm \sqrt{b^2−4ac}}{2a} \tag{15S.8}\]

    This equation, known as the quadratic formula, has two roots:

    \[x= \dfrac{−b + \sqrt{b^2−4ac}}{2a} \tag{15S.9}\]
    \[x= \dfrac{−b - \sqrt{b^2−4ac}}{2a} \tag{15S.10}\]

    Thus we can obtain the solutions to a quadratic equation by substituting the values of the coefficients (\(a\), \(b\), \(c\)) into the quadratic formula.

    When you apply the quadratic formula to obtain solutions to a quadratic equation, it is important to remember that one of the two solutions may not make sense or neither may make sense. There may be times, for example, when a negative solution is not reasonable or when both solutions require that a square root be taken of a negative number. In such cases, we simply discard any solution that is unreasonable and only report a solution that is reasonable. Exercise 15S.1 gives you practice using the quadratic formula.

    It is far easier to numerically solve cubic equations rather than to manipulate these unwieldy solutions.

    15.S: The Quadratic Formula is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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