Quadratic Formula

Definition

1.In the entry on quadratic equations, we showed that an equation of this type, written in standard form $ax^2 + bx + c = 0,$ involves a polynomial of degree two and always has exactly two roots in $\mathbb{C},$ counted with multiplicity, by the fundamental theorem of algebra. As we will recall later in this entry, the nature of the roots is determined by the sign of the discriminant $\Delta = b^2 - 4ac.$ The roots are most often found using the quadratic formula:

$$x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \tag{1}$$

2.Formula $(1)$ provides a straightforward procedure for solving any quadratic equation. It applies to equations with real or complex coefficients, provided that $a \neq 0,$ since $a$ appears in the denominator. The following points describe the terms in $(1):$

4.Formula $(1)$ also allows us to study how the roots vary when the coefficients depend on a parameter, as shown in the entry on quadratic equations with parameters.

5.For the following classification and its geometric interpretation, we assume that the coefficients $a,$ $b$ and $c$ are real.

6.As noted at the beginning, the expression under the square root, $b^2 - 4ac,$ is called the discriminant. Its sign uniquely determines the number and nature of the solutions.

7.When $\Delta > 0,$ the equation has two distinct real solutions. We write the solution set as $S = \{\ x_1, x_2 \ \}$ with $x_1, x_2 \in \mathbb{R}$ and $x_1 \neq x_2.$ The solutions are obtained by applying $(1)$ directly.

8.When $\Delta = 0,$ the two real roots coincide, so the equation has a single root of multiplicity two. We write the solution set as $S = \{\ x \ \}$ with $x \in \mathbb{R}$ and $x = x_1 = x_2.$ Since the discriminant is zero, $(1)$ reduces to the following formula for the solution.

$$x = -\frac{b}{2a} \tag{2}$$

9.Finally, when $\Delta < 0,$ the equation has no real solutions but always has two complex conjugate solutions with nonzero imaginary parts, which we express as $\nexists\ x \in \mathbb{R}.$ The complex solutions are obtained from $(1)$ by writing the square root as a multiple of the imaginary unit $i:$

$$x_{1,2} = \frac{-b \pm i\sqrt{4ac - b^2}}{2a}\tag{3}$$

10.The discriminant also determines the position of the graph of the quadratic function $f(x) = ax^2 + bx + c$ relative to the $x$-axis. Geometrically, a quadratic equation is associated with a parabola whose intersections with the axis depend on the sign of the discriminant:

11.More precisely:

Proof

13.We derive $(1)$ from the standard form of a quadratic equation by completing the square. We first rewrite $ax^2 + bx + c = 0$ with the constant term on its own on the right-hand side:

$$ax^2 + bx = -c$$

14.To complete the square, we divide both sides by $a,$ which is nonzero by assumption, and obtain:

$$x^2 + \frac{b}{a}x = -\frac{c}{a} \tag{4}$$

15.We now turn the left-hand side into a perfect square. The square of a binomial, $(a+b)^2,$ is equal to $a^2 + 2ab + b^2.$ To make the left-hand side of $(4)$ a perfect square, we therefore need to add the term corresponding to $b^2$ in this identity. We rewrite $(4)$ as follows:

$$x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} + \left(\frac{b}{2a}\right)^2$$

16.The left-hand side is now the square of a binomial. Writing it in factored form and simplifying the right-hand side gives:

$$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2} \tag{5}$$

17.We take square roots, including both signs because a square root and its negative have the same square. We can therefore rewrite $(5)$ as follows:

$$x + \frac{b}{2a} = \pm\frac{\sqrt{b^2 - 4ac}}{2a}$$

18.Isolating $x$ gives $(1):$

$$x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

19.After applying the quadratic formula, we can also check the solutions by comparing their sum and product with the coefficients of the equation. Vieta's formulas provide this check. They state that, for a quadratic equation $ax^2 + bx + c = 0$ with roots $x_1$ and $x_2,$ the following relations hold:

$$ \begin{align} x_1 + x_2 &= -\frac{b}{a} \\[6pt] x_1x_2 &= \frac{c}{a} \end{align} $$

20.These relations hold in $\mathbb{C}$ for any value of the discriminant. They follow by expanding the factored form $a(x - x_1)(x - x_2)$ and comparing coefficients.

Examples

21.We begin by applying $(1)$ to the first equation, $x^2 - 4x + 2 = 0.$ The equation is already in standard form, with $a = 1,$ $b = -4$ and $c = 2,$ so substituting these coefficients into the formula gives:

$$ \begin{align*} x_{1,2} &= \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(2)}}{2(1)} \\[6pt] &= \frac{4 \pm \sqrt{16 - 8}}{2} \\[6pt] &= \frac{4 \pm \sqrt{8}}{2} \\[6pt] &=\frac{4 \pm 2\sqrt{2}}{2} \end{align*} $$

22.Since $\Delta = 8 > 0,$ the equation has two distinct real solutions, namely $x_1 = 2 - \sqrt{2}$ and $x_2 = 2 + \sqrt{2}.$

23.The second equation requires a few steps to put it into standard form:

$$\frac{(x-1)^2}{2} - \frac{(x+1)(x-2)}{3} = \frac{x-1}{3}$$

24.We first clear the denominators and expand the square and the products, obtaining:

$$ \begin{align} 3(x^2 - 2x + 1) - 2(x^2 - x - 2) &= 2x - 2 \\[6pt] 3x^2 - 6x + 3 - 2x^2 + 2x + 4 &= 2x - 2 \\[6pt] x^2 - 4x + 7 &= 2x - 2\\[6pt] x^2 - 4x + 7 - 2x + 2 &= 0 \\[6pt] x^2 - 6x + 9 &= 0 \end{align} $$

25.The original equation is now in standard form. Its coefficients are $a = 1,$ $b = -6$ and $c = 9,$ and the discriminant is:

$$\Delta = (-6)^2 - 4(1)(9) = 36 - 36 = 0$$

26.Since $\Delta = 0,$ the two roots coincide. Substituting the coefficients into $(2),$ which, as noted earlier, is obtained from $(1)$ when the discriminant is zero, gives:

$$x = -\frac{b}{2a} = -\frac{-6}{2(1)} = 3$$

27.The equation therefore has a single real root $x = 3,$ of multiplicity two.

28.Finally, we solve $x^2 + 2x + 5 = 0.$ The coefficients are $a = 1,$ $b = 2$ and $c = 5,$ and this time the discriminant is negative:

$$\Delta = 2^2 - 4(1)(5) = 4 - 20 = -16$$

29.The equation therefore has no real solutions. We express the square root using the imaginary unit $i,$ so that $\sqrt{-16} = 4i,$ and obtain from $(3):$

$$x_{1,2} = \frac{-2 \pm 4i}{2} = -1 \pm 2i$$

30.The two solutions are the complex conjugates $x_1 = -1 - 2i$ and $x_2 = -1 + 2i.$