Using the Quadratic Formula Examples

Numerical Methods for Roots of Polynomials - Part II

J.M. McNamee , V.Y. Pan , in Studies in Computational Mathematics, 2013

12.4 Errors in the Quadratic Solution

The quadratic formula seems very simple in theory, but in actual calculations on a computer there can be many pitfalls, as discussed for example by Forsythe (1969, 1970). Problems are caused by round-off error in floating-point arithmetic, and even more so by the possibility of overflow or underflow. Forsythe considers the set of normalized floating-point numbers, and gives detailed specifications for a satisfactory solver. He points out that many computer systems halt a program on overflow, and/or set the result to 0 on underflow. In such cases, the programmer must take great pains to ensure that over- or underflow can never occur. Forsythe gives several numerical examples where over- or underflow or badly erroneous results can occur, unless precautions are taken. He points out that quadratic equations occur as sub-problems in the numerical solution of general polynomial equations by Muller's or Laguerre's method, so that a good quadratic solver is a necessity. He mentions a Fortran IV algorithm by Kahan which meets his specifications, but this is probably inaccessible today.

In his (1970) paper Forsythe considers the problem of cancelation of nearly equal numbers. This occurs in the standard formula (with the positive sign on the square root) if $b^2\gg 4\mathit{ac}$ and $b>0$ , or with the negative square root if $b<0$ . The cure for $b>0$ is to use

(12.17) ${\zeta }_1=\frac{2c}{-b-\sqrt{b^2-4\mathit{ac}}}$

with

(12.18) ${\zeta }_2=\frac{-b-\sqrt{b^2-4\mathit{ac}}}{2a}$

while for $b<0$ we may use

(12.19) ${\zeta }_1=\frac{-b+\sqrt{b^2-4\mathit{ac}}}{2a}$

and

(12.20) ${\zeta }_2=\frac{2c}{-b+\sqrt{b^2-4\mathit{ac}}}$

Finally, special precautions must be taken if a is equal or close to 0, as may happen for example in Muller's method as we approach a root. The cases a = b = c = 0, or a = b = 0 and $c\ne 0$ also deserve special consideration.

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Analytic Geometry

S.M. Blinder , in Guide to Essential Math (Second Edition), 2013

5.3 Conic Sections

Mathematically, a right-circular cone is a surface swept out by a straight line, with one point—the vertex—kept fixed while the line sweeps around a circular path. You would have to place two ice-cream cones point-to-point to simulate a mathematical cone, as shown in Figure 5.3. Greek mathematicians (Apollonius is usually credited) discovered that planes intersecting the cone at different angles produce several interesting curves, which are called conic sections. Degenerate cases, in which the plane passes through the vertex, give either a single point, a straight line, or two intersecting lines. The nondegenerate conic sections, illustrated in Figure 5.4, are circles, ellipses, parabolas, and hyperbolas.

Figure 5.3. Mathematical definition of a cone and one possible physical realization. Each of the two sheets on opposite sides of the vertex is called a nappe.

Figure 5.4. Intersections of the cone and plane showing how the different conic sections are generated.

A right-circular cone can be represented by the three-dimensional equation:

(5.17) $x^2+y^2=z^2\text{.}$

For each value $\pm z$ , this corresponds to a circle of radius z. Also, by analogy with Eq. (5.7) for a straight line, we surmise that the equation for a plane in three dimensions has the form

(5.18) $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\text{.}$

The intersection of the cone with an arbitrary plane is satisfied by points $(x\text{,}y\text{,}z)$ which are simultaneous solutions of Eqs. (5.17) and (5.18). The equation for the surface of intersection, in the form $f(x\text{,}y)=0$ , can be obtained by eliminating z between the two equations. This implies that a conic section has the general form

(5.19) ${\mathit{Ax}}^2+\mathit{Bxy}+{\mathit{Cy}}^2+\mathit{Dx}+\mathit{Ey}+F=0\text{.}$

Assuming it is nondegenerate, the form of the conic section is determined by its discriminant, $B^2-4\mathit{AC}$ . If the discriminant is positive, it is a hyperbola, if it is negative, an ellipse or a circle. A discriminant of zero implies a parabola. The coefficients $D\text{,}E$ , and F help determine the location and scale of the figure.

The simplest nondegenerate conic section is the circle. A circle centered at $(x_0\text{,}{y}_0)$ with a radius of a satisfies the equation

(5.20) $(x-x_0{)}^2+(y-y_0{)}^2=a^2\text{.}$

The unit circle is the special case when $x^2+y^2=1$ . A circle viewed from an angle has the apparent shape of an ellipse. An ellipse centered at $(x_0\text{,}{y}_0)$ with semimajor axis a and semiminor axis b, as shown in Figure 5.5, is described by the equation

Figure 5.5. Ellipse with semimajor axis $a$ and semiminor axis $b$ .

(5.21) $\frac{(x-x_0{)}^2}{a^2}+\frac{(y-y_0{)}^2}{b^2}=1\text{.}$

It is assumed that $a>b$ , so that the long axis of the ellipse is oriented in the x-direction. When $b=a$ , the ellipse degenerates into a circle.

Figure 5.6 represents a hyperbola with the equation:

Figure 5.6. Hyperbola with semimajor axis $a$ and semiminor axis $b$ . Asymptotes are shown in gray.

(5.22) $\frac{(x-x_0{)}^2}{a^2}-\frac{(y-y_0{)}^2}{b^2}=1\text{.}$

The hyperbola can likewise be characterized by a semimajor axis a and a semiminor axis b, which, in this case, define a rectangle centered about $(x_0\text{,}{y}_0)$ . The two branches of the hyperbola are tangent to the rectangle, as shown. A distinctive feature are the asymptotes, the two diagonals of the rectangle extended to infinity. Their equations are

(5.23) $(y-y_0)=\pm \sqrt{\frac{b}{a}}(x-x_0)\text{.}$

When $b=a$ , the asymptotes become perpendicular, and we obtain what is called an equiangular or rectangular hyperbola. A simple example is the unit hyperbola

(5.24) $x^2-y^2=1\text{.}$

Recalling that $x^2-y^2=(x+y)(x-y)$ , we can make a transformation of coordinates in which $(x+y)\to x$ and $(x-y)\to y$ . This simplifies the equation to

(5.25) $\mathit{xy}=1\text{.}$

The hyperbola has been rotated by 45° and the asymptotes have become the coordinate axes themselves. These rectangular hyperbolas are shown in Figure 5.7. More generally, hyperbolas of the form $\mathit{xy}=\mathrm{const}$ represent relations in which y is inversely proportional to x. An important example is Boyle's law relating the pressure and volume of an ideal gas at constant temperature, which can be expressed as $\mathit{pV}=\mathrm{const}$ .

Figure 5.7. Rectangular hyperbolas.

We have already considered parabolas of the form $(y-y_0)=k(x-x_0{)}^2$ , in connection with the quadratic formula. Analogous "sideways" parabolas can be obtained when the roles of x and y are reversed. Parabolas, as well as ellipses and hyperbolas, can be oriented obliquely to the axes by appropriate choices of B, the coefficient of xy, in the conic-section equation (5.19), the simplest example being the $45°$ hyperbola $\mathit{xy}=1$ considered above. Parabolas have the unique property that parallel rays incident upon them are reflected to a single point, called the focus, as shown in Figure 5.8. A parabola with the equation

Figure 5.8. Left: Focusing of parallel lines by a parabola. Right: Paraboloid dish antenna.

(5.26) $y^2=4\mathit{px}$

has its focus at the point $(p\text{,}0)$ . A parabola rotated about its symmetry axis generates a paraboloid. This geometry is exploited in applications where radiation needs to be concentrated at one point, such as radio telescopes, television dishes, and solar radiation collectors; also, when light emitted from a single point is to be projected as a parallel beam, as in automobile headlight reflectors.

Problem 5.3.1

Find the center and radius of the circle with equation ${\mathit{Ax}}^2+{\mathit{Ay}}^2+\mathit{Bx}+\mathit{Cy}+D=0$ .

Problem 5.3.2

Propose equations of three full circles and two semicircles which can be used to construct a Yin-Yang symbol.

Problem 5.3.3

The latus rectum (pardon the expression!) of a conic section is the length of a chord parallel to the directrix. Find the latus rectum of the ellipse $\frac{(x-x_0{)}^2}{a^2}+\frac{(y-y_0{)}^2}{b^2}=1$ and of the parabola $y^2=4\mathit{px}$ .

Problem 5.3.4

Find the four points of intersection of the confocal ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ .

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Algebra, Abstract

KiHang Kim , Fred W. Roush , in Encyclopedia of Physical Science and Technology (Third Edition), 2003

VI.C Applications to Solvability and Constructibility

The problem of solving quadratic equations goes at least back to the Babylonians. In the ninth century, the Muslim mathematician Al-Khwarismi gave a version of the modern quadratic formula. In the mid-sixteenth century, Italian mathematicians reduced the solution of a cubic equation to the form

${\text{x}}^3+\text{m}\text{x}=\text{n}$

by dividing it by the coefficient of x ³ and making a substitution replacing x by some x  − c. They then solved it as

$\begin{array}{c}\hfill \text{x}=\text{a}-\text{b}\hfill \\ \text{a}=\sqrt[3]{(\text{n}/2)+\sqrt{{(\text{n}/2)}^2+{(\text{m}/3)}^3}}\hfill \\ \text{b}=\sqrt[3]{-(\text{n}/2)+\sqrt{{(\text{n}/2)}^2+{(\text{m}/3)}^3.}}\hfill \end{array}$

They proceeded to solve quartic equations by reducing them to cubics. But no one was able to solve the general quintic using nth roots and in 1824 N.H. Abel proved this is impossible. E. Galois in his brief life proved this also, as part of a general theory that applies to all polynomials. This is based on field theory, and we describe it next.

In the extensions F(t), one root of a polynomial p(t) has been added, or adjoined, to F. Extensions obtained by adding all roots of a polynomial are called normal extensions. The roots can be added one at a time in any order.

Finite dimensional normal extensions can be studied by finite groups called Galois groups. The Galois group of a normal extension F  ⊂ E is the group of all field automorphisms of E that are the identity on F. It will, in effect, permute the roots of a polynomial whose roots generate the extension. For example, let F  = Q(ξ) and let E  = F( $\sqrt[3]{2}$ ), where $\xi =(-1+\text{i}\sqrt{3})/2$ . Then an automorphism of E exists taking $\sqrt[3]{2}$   →   ξ $\sqrt[3]{2}$ ξ $\sqrt[3]{2}$   →   ξ²( $\sqrt[3]{2}$ ), ξ²( $\sqrt[3]{2}$ )→ $\sqrt[3]{2}$ . The Galois group is cyclic of order 3, generated by this automorphism. Since the ratio ξ of two roots goes to itself, it is the identity on Q(ξ).

The order of the Galois group equals the degree of a normal extension. Moreover, there is a 1–1 correspondence between subfields F  ⊂ K  ⊂ E and subgroups of H  ⊂ G, the Galois group of E over F. To a subgroup H is associated the field k  =   {x  ∈ E  : f(x)   = x for all f  ∈ K}.

A splitting field of a polynomial p over a field F is a minimal extension of F over which p factors into factors of degree 1. It is a normal extension and any two splitting fields are isomorphic.

Suppose a polynomial p is solvable by radicals over Q. Let E be the splitting field of p over Q. Each time we extract a radical the roots of the radical generate a normal extension F ₁ of a previous field F ₂. Let E _i   = F _i ∩ E. Then F ₂ over F ₁ has cyclic Galois group, so E ₂ over E ₁ does also.

It follows that there exist a series of extensions Q  = D ₀  ⊂ D ₁  ⊂   ⋯   ⊂ D _n   = E each normal over the preceding such that the Galois group of each over the other is cyclic. It follows that the Galois group G has a series of subgroups G_n   =   {e}   ⊂ G _n−1  ⊂   ⋯   ⊂ G ₀  = G such that g_i is a normal subroup of G _i−1 with cyclic quotient group. Such a group is called solvable.

The symmetric group of degree 5 has as its only nontrivial proper normal subgroup the alternating group which is simple. Therefore, it is not solvable. If F(x) is a degree 5 irreducible polynomial over Q with exactly two nonreal roots, there exists an order 5 element in its Galois group just because 5 divides the degree of the splitting field. Complex conjugation gives a transposition. Therefore, the Galois group is $\mathcal{L}$ ₅. So polynomials of degree 5 cannot in general be solved by radicals.

Conversely, it is true that every normal extension E  ⊂ F with cyclic Galois group can be generated by radicals. It can be shown that there is a single element θ such that E  = F(θ) (consider all linear combinations θ of a basis for E over F, and there being a finite number of intermediate fields).

Let the extension be cyclic of order n and let τ be such that τⁿ  =   1 but no lower power. Let the automorphism g generate the Galois group. Let t  =   θ   +   τg(θ)   +   ⋯   +   τ^n-1 g^n-1(θ). Then t has n distinct conjugates (assuming τu   ∈ F) g ⁱ(θ)   +   τ gⁱ⁺¹(θ)   +   ⋯   +   τ^n-1g^n-1+i(θ) and so its minimum polynomial has degree n. Since g(t)   =   τ^{− 1}(t), the element t ⁿ  = a is invariant under the Galois group and lies in F. So θa, g(θ),…,g^n-1(θ) lie in the splitting field of x ⁿ  = a, which must be E.

Geometric constructions provide an application of field theory. Suppose we are given a unit line segment. What figures can be constructed from it by ruler and compass? Let the segment be taken as a unit length or the x axis. Wherever we construct a new point from existing ones by ruler and compass it is an intersection of a line or circle with a line or circle. Such intersections lead to quadratic equations. Therefore, if a point P is constructible, each coordinate must be obtained from rational numbers by adding, subtracting, multiplying, dividing, or taking square roots. Such quantities lie in an extension field of E  ⊂ Q such that there exist fields E ₀  = Q  ⊂ E ₁  ⊂   ⋯   ⊂ E _k   = E and ${\mathbf{E}}_{\mathbf{n}}={\mathbf{E}}_{\mathbf{n}-1}\left(\sqrt{\text{a}}\right)$ for a  ∈ E _n−1. The degree of [E  : Q]   =   [E _n   : E _n−1]⋯[E ₁  : E ₀] is a power of 2.

Therefore, if x is a coordinate of a constructible point, x lies in an extension of degree 2 ⁿ , in fact a normal extension of degree 2 ⁿ . But if [Q(x)   : Q] has degree not a power of 2, this is impossible since [E  : Q]   =   [E  : Q(x)] [Q(x)   : Q].

In particular, duplicating a cube (providing a cube of volume precisely 2) and trisecting an angle of 60° lead to roots of irreducible cubics x ³  −   2   =   0 and 4cos³θ−3cosθ−cos 60°   =   0 and cannot be performed. Since π i does not satisfy any monic polynomial with coefficients in Q, the circle cannot be squared.

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Equations, Inequalities, and Modeling

Sarhan M. Musa , in Fundamentals of Technical Mathematics, 2016

3.1.3 Quadratic equations

Definition of quadratic equation

A quadratic equation is a second order equation written as ax ²  + bx  + c  =   0 where a, b, and c are coefficients of real numbers and a  ≠   0.

Definition of quadratic formula

The quadratic formula is a general formula used for solving the quadratic equation:

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

Note that: if x ²  = k, where k  ≥   0, then $x=\sqrt{k}$ or $-\sqrt{x}$ same as $x=\pm \sqrt{k}$ , where ± means "plus or minus."

Example 1

Solve the following equations:

(a)

2x ²  +   3x  −   4   =   0

(b)

x ²  +   5x  −   3   =   0

Solution

(a)

2x ²  +   3x  −   4   =   0

The coefficients of the equations are:

a  =   2, b  =   3, c  =   −4

So, $x=\frac{-3\pm \sqrt{3^2-4·2·\left(-4\right)}}{2·2}=\frac{-3\pm \sqrt{9+32}}{4}=\frac{-3\pm \sqrt{41}}{4}$

(b)

x ²  +   5x  −   3   =   0

The coefficients of the equations are:

a  =   1, b  =   5, c  =   −3.

So, $x=\frac{-5\pm \sqrt{5^2-4·1·\left(-3\right)}}{2·1}=\frac{-5\pm \sqrt{25+12}}{2}=\frac{-5\pm \sqrt{37}}{2}\text{.}$

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Complex Numbers and Functions

Frank E. Harris , in Mathematics for Physical Science and Engineering, 2014

Inverse Circular and Hyperbolic Functions

You probably already know that the inverse circular functions are multiple-valued; this must be so because many angles have the same value of the sine or cosine.

If we multiply the $\sin \theta$ formula of Eq. (3.28) by $e^{i\theta }$ and rearrange it to

${\left(e^{i\theta }\right)}^2-2i\sin \theta \left(e^{i\theta }\right)-1=0\text{,}$

we can identify it as a quadratic equation in $e^{i\theta }$ which we can then solve, using the quadratic formula, to reach

(3.42) $e^{i\theta }=i\sin \theta +\sqrt{-{\sin }^2\theta +1}\text{.}$

We have chosen to take the plus sign for the radical.

If we now set $\sin \theta =z$ and $\theta =\arcsin z={\sin }^{-1}z$ , and then take the logarithm of both sides of Eq. (3.42), we will have succeeded in inverting the equation for $\sin \theta$ :

${\sin }^{-1}z=-i\ln \left[\mathit{iz}+\sqrt{1-z^2}\right]\text{.}$

The appearance of the logarithm causes this formula to be multiple-valued; if we add $2\pi i$ to the logarithm we see that it increases ${\sin }^{-1}x$ by $2\pi$ , an expected result because the sine has period $2\pi$ . Within each interval of length $2\pi$ there are two angles with the same value of the sine; that not given above is obtained by taking the minus sign for the radical in the quadratic formula.

Similar analyses can be carried out for all the circular and hyperbolic functions. Some such results, in forms consistent with the usual designations of principal value ( $-\pi /2\le {\sin }^{-1}x\le \pi /2$ , $-\pi /2<{\tan }^{-1}x<\pi /2$ ), and with all functions real for real arguments, are

$\begin{array}{cc}{\sin }^{-1}z=-i\ln \left[\mathit{iz}+\sqrt{1-z^2}\right]\text{,}\hfill & {\sinh }^{-1}z=\ln \left[z+\sqrt{1+z^2}\right]\text{,}\hfill \\ {\tan }^{-1}z=\frac{i}{2}\ln \left(\frac{1-\mathit{iz}}{1+\mathit{iz}}\right)\text{,}\hfill & {\tanh }^{-1}z=\frac{1}{2}\ln \left(\frac{1+z}{1-z}\right)\text{.}\hfill \end{array}$

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Problem Solving and the Solution of Algebraic Equations

Robert G. Mortimer , in Mathematics for Physical Chemistry (Fourth Edition), 2013

5.1.1 Polynomial Equations

A polynomial equation is written in the form

(5.1) $f(x)=a_0+a_{1}x+a_2{x}^2+\cdots +a_n{x}^n=0\text{,}$

where the as are constants. The integer n is called the degree of the equation. If n  =   1, the equation is a linear equation. If n  =   2, the equation is a quadratic equation. If n  =   3, the equation is a cubic equation. If n  =   4, it is a quartic equation, and so on. Generally, there are n roots to an nth-degree polynomial equation, but two or more of the roots can be equal to each other. For most equations arising from chemical problems, there will be only one root that is physically reasonable, and the others must be disregarded. For example, a concentration cannot be negative, and if a quadratic equation for a concentration produces a positive root and a negative root, the negative root must be disregarded. It is also possible for some of the roots to be imaginary or complex numbers. Complex roots cannot represent physically measurable quantities and must be disregarded if we are solving for a physically meaningful quantity.

Linear Equations

A linear equation is of the form

(5.2) $a_0+a_{1}x=0\text{.}$

This equation has a single root:

(5.3) $x=-\frac{a_0}{a_1}\text{.}$

Quadratic Equations

A quadratic equation can be written in a standard form as

(5.4) ${\mathit{ax}}^2+\mathit{bx}+c=0\text{.}$

A quadratic equation generally has two roots, which can be equal to each other. Some quadratic expressions can be factored, which means that the equation can be written

(5.5) $a(x-x_1)(x-x_2)=0\text{,}$

where x ₁ and x ₂ are the two roots of the equation. If a quadratic equation cannot easily be factored, you can apply the quadratic formula

(5.6)

The quadratic formula delivers two roots, one when the positive sign in front of the square root is chosen and the other when the negative sign is chosen. There are three cases:

1.

if the discriminant $b^2-4\mathit{ac}$ is positive, the roots will be real and unequal;

2.

if the discriminant is equal to zero, the two roots will be real and equal to each other;

3.

if the discriminant is negative, the roots will be complex and each root is the complex conjugate of the other root.

Exercise 5.1

Show by substitution that the quadratic formula provides the roots to a quadratic equation.

A common application of a quadratic equation in elementary chemistry is the calculation of the hydrogen-ion concentration in a solution of a weak acid. If activity coefficients are assumed to equal unity, the equilibrium expression in terms of molar concentrations is

(5.7) $K_{\mathrm{a}}=\frac{([{\text{H}}^{+}]/\text{c}°)([{\text{A}}^{-}]/\text{c}°)}{[\text{HA}]/\text{c}°}\text{,}$

here [H⁺] represents the hydrogen-ion concentration expressed in mol   l⁻¹ (molarity), [A⁻] represents the acid-anion concentration, [HA] represents the concentration of the undissociated acid, the constant c° is defined to equal exactly 1   mol   l⁻¹, and $K_{\mathrm{a}}$ represents the acid ionization constant. It is true that the hydrogen ions are nearly all attached to water molecules or water molecule dimers, and so on, so that we could write [H₃O⁺] instead of [H⁺], but this makes no difference in the calculation. The expression in terms of molalities (moles per kilogram of solvent)can also be used and has the same appearance.

Example 5.1

For acetic acid, $K_{\mathrm{a}}=1.754\times {10}^{-5}$ at 25   °C. Find [H⁺] if 0.1000   mol of acetic acid is dissolved in enough water to make 1.000   l of solution. We say that the stoichiometric concentration (the concentration that would occur if there were no ionization) is equal to 0.100   mol   l⁻¹. Assume that activity coefficients are equal to unity.

If no other sources of hydrogen ions or acetate ions are present, [H⁺]/c°   =   [A⁻]/c°, which we denote by x,

$K_{\mathrm{a}}=\frac{x^2}{0.1000-x}$

or

$x^2+K_{\mathrm{a}}x-0.1000K_{\mathrm{a}}=0\text{.}$

From Eq. (5.6), our solution is

$\begin{array}{cc}\hfill x=& \frac{-K_{\mathrm{a}}\pm \sqrt{K_{\mathrm{a}}^2+0.4000K_{\mathrm{a}}}}{2}\hfill \\ \hfill =& 1.316\times {10}^{-3}\text{or}-1.333\times {10}^{-3}\text{,}\hfill \\ \hfill [{\text{H}}^{+}]=& [{\text{A}}^{-}]=1.316\times {10}^{-3}\text{mol}{\text{l}}^{-1}\text{.}\hfill \end{array}$

We disregard the negative root because a concentration cannot be negative.

Exercise 5.2

For hydrocyanic acid (HCN), $K_{\mathrm{a}}=4.9\times {10}^{-10}$ at 25   °C. Find [H⁺] if 0.1000   mol of hydrocyanic acid is dissolved in enough water to make 1.000   l. Assume that activity coefficients are equal to unity and neglect hydrogen ions from water.

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Functions

Huw Fox , Bill Bolton , in Mathematics for Engineers and Technologists, 2002

The quadratic formula

Consider the quadratic equation ax ² + bx + c = 0. To obtain the roots by completing the square method, we divide throughout by a to give:

$x^2+\frac{b}{a}x+\frac{c}{a}=0$

This can be written as:

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

To make the left-hand side of the equation a perfect square we must add (b/2a)² to both sides of the equation. Hence:

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

and so:

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

Taking the square root of both sides of the equation gives

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

Thus:

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

and so we have the general formula for the solution of a quadratic equation:

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

Consider the following three situations:

•

If we have (b ² − 4ac) > 0, then the square root is of a positive number. There are then two distinct roots which are said to be real.

•

If we have (b ² − 4ac) = 0, then the square root is zero and the formula gives just one value for x. Since a quadratic equation must have two roots, we say that the equation has two coincident real roots.

•

If we have (b ² − 4ac) < 0, then the square root is of a negative number. A new type of number has to be invented to enable such expressions to be solved. The number is referred to as a complex number and the roots are said to be imaginary (the roots in 1 and 2 above are said to be real). Such numbers are discussed later in this book.

Key point

The general formula for the solution of a quadratic equations is:

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

Example

Determine, if they exist as real roots, the roots of the following quadratic equations:

(a)

4x ² − 7x + 3,

(b)

x ² − 4x + 4,

(c)

x ² + 2x + 4.

(a)

Using the general quadratic formula [ 10], here we have a = 4, b = −7 and c = 3. Therefore:

$x=\frac{+7\pm \sqrt{49-48}}{8}=\frac{7\pm 1}{8}$

Therefore, x = 1 or x = 0.75 defines the two roots of the equation. We can now represent the function as:

$\left(x-1\right)\left(x-0.75\right)$

and this, when multiplied out, gives x ² − 1.75x + 0.76 and which when multiplied by 4 gives the original equation 4 x ² − 7x + 3.

(b)

Using the general quadratic formula [10] gives:

$x=\frac{-2\pm \sqrt{16-16}}{2}=2$

Therefore, we have two roots with x = 2. We may now rewrite the equation in the form (x − 2)(x − 2). This, when multiplied out, gives x ² − 4x + 4.

(c)

Using the general quadratic formula [10] gives:

$x=\frac{-2\pm \sqrt{4-16}}{2}$

Since the term inside the square root is negative, we have no real roots.

Example

Figure 1.13 shows a simple cantilever of length L, propped at its free end. It can be shown that the bending moment M of this type of cantilever is a function of the distance x measured from the fixed end of the beam, thus M = f(x). The defining equation for the function is:

$M=f\left(x\right)=\frac{W{x}^2}{2}-\frac{5WLx}{8}+\frac{W{L}^2}{8}$

where W is the distributed load in newtons per unit length. Using this quadratic formula, determine the positions along the beam at which the bending moment is zero (in engineering called the points of contraflexure).

Figure 1.13. Propped cantilever

When M = 0, we have:

$\frac{W{x}^2}{2}-\frac{5WLx}{8}+\frac{W{L}^2}{8}=0$

We can solve this by using the general quadratic formula [10]. Firstly, we can simplify the expression by multiplying through by 8 and taking the W term out as a factor:

$4W{x}^2-5WLx+W{L}^2=W\left(4x^2-5Lx+L^2\right)=0$

and so the equation becomes 4x ² − 5Lx + L ² = 0. Thus:

$\begin{array}{l}x=\frac{5L\pm \sqrt{{\left(-5L\right)}^2-4\left(4\right)\left(L^2\right)}}{2\left(4\right)}\hfill \\ x=\frac{5L\pm \sqrt{25L^2-16L^2}}{8}\hfill \end{array}$

Hence x = L/4 or x = L. The bending moment is thus zero at two locations (it has two points of contraflexure), i.e. at L/4 from the fixed end or at the extreme right-hand end of the beam when x = L.

Example

The distance s in metres moved by a vehicle over a period of time t seconds is defined by the equation s = ut + ½at ², with a being the constant acceleration and u the initial velocity. Assuming the vehicle commences motion with an initial velocity of 5 m/s and covers 84 m with a constant acceleration of 2 m/s², calculate the time over which this occurs.

Substituting the values in the equation gives:

$\begin{array}{l}84=5t+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.(2)t^2\hfill \\ 84=5t+t^2\hfill \end{array}$

Writing the equation in the general format:

$t^2+5t-84=0$

Thus:

$t=\frac{-5\pm \sqrt{25-4\left(84\right)}}{2}$

and so t = −12 s or t = 7 s. Since we cannot have a negative time, the only acceptable answer is t = 7 s.

The solution may be checked by substituting into the original equation t ² + 5t − 84 = 0, when t = 7 s we have 7² + 5(7) − 84 = 0. Since this is true, our solution holds.

Example

The total surface area A of a cylinder of radius r and height h is given by the equation A = 2πr ² + 2πrh. If h = 6 cm, what will be the radius required to give a surface area of 88/7 cm²? Take π as 22/7.

Putting the numbers in the equation gives

$\frac{88}{7}=2\times \frac{22}{7}{r}^2+2\times \frac{22}{7}\times 6r$

Multiplying throughout by 7 and dividing by 44 gives

$2=r^2+6r$

Hence we can write

$r^2+6r-2=0$

and so:

$r=\frac{-6\pm \sqrt{36-4\times 1\times \left(-2\right)}}{2\times 1}=\frac{-6\pm \sqrt{44}}{2}=\frac{-6\pm 6.63}{2}$

Hence the solutions are r = −6.32 cm and r = 0.32 cm. The negative solution has no physical significance. Hence the solution is a radius of 0.32 cm.

We can check this value of 0.32 cm by substitution in the equation 2 = r ² + 6r. Hence 0.10 + 1.92 = 2.02, which is effectively 2 bearing in mind the rounding of the root value to two decimal places that has occurred.

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Complex Vector Spaces and General Inner Products

Stephen Andrilli , David Hecker , in Elementary Linear Algebra (Fifth Edition), 2016

Diagonalizable Complex Matrices and Algebraic Multiplicity

We say a complex matrix A is diagonalizable if and only if there is a nonsingular complex matrix P such that P ⁻¹ AP = D is a diagonal matrix. Just as with real matrices, the matrix P has fundamental eigenvectors for A as its columns, and the diagonal matrix D has the eigenvalues for A on its main diagonal, with d _ii being an eigenvalue corresponding to the fundamental eigenvector that is the ith column of P. The six-step Method for diagonalizing a matrix given in Section 3.4 works just as well for complex matrices.

The algebraic multiplicity of an eigenvalue of a complex matrix is defined just as for real matrices — that is, k is the algebraic multiplicity of an eigenvalue λ for a matrix A if and only if (x−λ) ^k is the highest power of (x − λ) that divides p _A (x). However, an important property of complex polynomials makes the situation for complex matrices a bit different than for real matrices. In particular, the Fundamental Theorem of Algebra states that any complex polynomial of degree n factors into a product of n linear factors. Thus, for every n × n matrix A, p _A (x) can be expressed as a product of n linear factors. Therefore, the algebraic multiplicities of the eigenvalues of A must add up to n. This eliminates one of the two reasons that some real matrices are not diagonalizable. However, there are still some complex matrices that are not diagonalizable, as we will see later in Example 4.

Example 3

Consider the matrix

$\mathbf{A}=\left[\begin{array}{ll}cos\theta & -sin\theta \\ sin\theta & cos\theta \end{array}\right]$

from Example 3.4 in Section 3.4 for a fixed value of θ such that $sin\theta \ne 0.$ In that example, we computed $p_{\mathbf{A}}(x)=x^2-2(cos\theta )x+1,$ which factors into complex linear factors as $p_{\mathbf{A}}(x)=(x-(cos\theta +isin\theta ))(x-(cos\theta -isin\theta )).$ Thus, the two complex eigenvalues ¹ for A are ${\lambda }_1=cos\theta +isin\theta$ and ${\lambda }_2=cos\theta -isin\theta .$

Row reducing λ ₁ I ₂ −A yields $\left[\begin{array}{ll}1& -i\\ 0& 0\end{array}\right],$ thus giving the fundamental eigenvector [i,1]. Similarly, row reducing λ ₂ I ₂ −A produces the fundamental eigenvector [−i,1]. Hence, $\mathbf{P}=\left[\begin{array}{ll}i& -i\\ 1& 1\end{array}\right].$ You can verify that

$\begin{array}{ll}\hfill {\mathbf{P}}^{-1}\mathbf{AP}& =\left(\frac{1}{2}\left[\begin{array}{ll}-i& 1\\ i& 1\end{array}\right]\right)\left[\begin{array}{ll}cos\theta & -sin\theta \\ sin\theta & cos\theta \end{array}\right]\left[\begin{array}{ll}i& -i\\ 1& 1\end{array}\right]\hfill \\ \hfill & =\left[\begin{array}{ll}\hfill cos\theta +isin\theta \hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill cos\theta -isin\theta \hfill \end{array}\right]=\mathbf{D}.\hfill \end{array}$

For example, if $\theta =\frac{\pi }{6},$ then $\mathbf{A}=\frac{1}{2}\left[\begin{array}{ll}\hfill \sqrt{3}\hfill & \hfill -1\hfill \\ \hfill 1\hfill & \hfill \sqrt{3}\hfill \end{array}\right]$ and $\mathbf{D}=\frac{1}{2}\left[\begin{array}{ll}\hfill \sqrt{3}+i\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \sqrt{3}-i\hfill \end{array}\right]$ . Note that the fundamental eigenvectors for A are independent of θ, and hence so is the matrix P. However, D and the eigenvalues of A change as θ changes.

This example illustrates how a real matrix could be diagonalizable when thought of as a complex matrix, even though it is not diagonalizable when considered as a real matrix.

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Complex numbers and some applications

Fatih Yılmaz , in Calculus for Engineering Students, 2020

3.1.1 Complex arithmetic

In elementary mathematics, complex numbers emerge under the conditions where simple algebraic equations have solutions that cannot be expressed in terms of real numbers alone. Although complex numbers occur in many branches of mathematics, they arise most directly when solving polynomial equations. Unfortunately not all equations have (real number) solutions. Consider the following general quadratic equation:

$a{x}^2+bx+c=0,$

where a, b, and c are real numbers. The quadratic formula

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

gives the zeros of the polynomial. It has no real solution for $\mathrm{\Delta }=b^2-4ac<0$ . For example,

$$\begin{gathered} x^2\ge 0;x\in \mathbb{R} \\ {x}^2+1\ge 0+1=1>0;x\in \mathbb{R} \\ {x}^2\ne -1;x\in \mathbb{R}. \end{gathered}$$

Around this problem, let us consider the function $f(x)=x^2+1$ , and the main question is where the function $f(x)$ crosses the x-axis. As it can be seen, it has no solution. However, about 200 years ago, Carl Friedrich Gauss proved that every polynomial of degree n has exactly n roots, which was later called the fundamental theorem of algebra. Taking into account this theorem, the function $f(x)=x^2+1=0$ must have two roots.

$f(x)=x^2+1$ Here, we introduce the symbol i, whose defining property is that it satisfies the equation

$i^2=-1.$

From its definition, it is clear that $i\notin \mathbb{R}$ . The notation i was first introduced by the Swiss mathematician and physicist Leonhard Euler (1707–1783).

A complex number is any symbol of the form $z=a+ib$ , or $z=a+bi$ , where a and b are any two real numbers. The real part of z is the number a, and the imaginary part of z is b; note that both parts are real numbers. One may define a complex number as nothing more than an ordered pair of two real numbers (a, b). The ordering is significant.

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Exploring Functions

S.J. Garrett , in Introduction to Actuarial and Financial Mathematical Methods, 2015

Roots

Another extremely useful concept is that of the roots of a function. A function f(x) is said to have a root at x = c if f(c) = 0. A function may have one root, many roots, or no roots in the domain of real numbers. You may find this phrasing rather awkward, but, as we shall see in Chapter 8, although a function may not have any real roots, it may have complex roots; that is, roots in the particular number system called complex numbers, $\mathbb{C}$ . This is stated without explanation here and we will continue to ignore complex roots until Chapter 8. Until that point, we will assume that everything is firmly based in terms of real numbers and the term "roots" should be understood as meaning "real roots."

As we can immediately see from the definition, the roots of a function are the values within the domain that solve the equation f(x) = 0. This concept provides a clear link between functions and the material of Section 1.3. Strategies that are used to find the roots of functions are identical to the strategies used to solve equations.

Example 2.3

Use an algebraic approach to find the roots of the function

$g(y)=y^2-4y+3$

Solution

The roots are the values of y such that

$y^2-4y+3=0$

Two strategies might immediately come to mind. The first is to use the standard quadratic formula. The second is to note that the LHS factorizes as ( y − 3)(y − 1). Both strategies lead to the correct result that g(y) has roots at y = 1 and y = 3.

Example 2.4

If

$g(x)=(x+1)(x^2-4x+3)$

find the values of x such that g(x) > 0.

Solution

The first stage in solving an inequality is to find the roots of the function. In this case, we note that the expression is already part factorized and so we can immediately see that x = −1 is a root. Other roots can be determined by noting that, in this case, the quadratic term is identical to that in Example 2.3. We then identify x = 1 and x = 3 as the remaining roots. We now need to determine the broader behavior of the function in order to solve the inequality. The function expands to

$f(x)=x^3-3x^2-x+3$

and we see its value at large values of x is dominated by x ³. The function is therefore negative for large negative values of x and positive for large positive values of x. With this information and knowledge that it crosses the horizontal axis three times at x = ±1 and x = 3, it should be clear that g(x) > 0 for x ∈ (−1,1) and $x\in (3,\infty )$ . We might opt to write these intervals as $x\in (-1,1)\cup (3,\infty )$ , but this is not essential to communicate the answer correctly.

At this stage in the book, it is assumed that your mathematical armory is not yet sufficient to attack root-finding problems in anything but simple polynomial functions. Where necessary, you should revert to graphical techniques, as discussed below, or software tools such as Wolfram Alpha or Excel's Goal Seek function. More advanced methods will be developed in later chapters.

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Using the Quadratic Formula Examples

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