Download Quadratic equations and complex numbers

Review for November Mathletes Contest:
Quadratic Equations and Complex Numbers
Written Saturday, 3 November 2001
Created for LSCHS Mathletes web page (http://www.angelfire.com/pa3/lschsmathletes). Please do not place on
other sites.
I have read this over carefully, but there may be small math errors in places. The theory is
correct.
1. Quadratic Equations and the Quadratic Formula
ax2 + bx + c = 0 is a quadratic equation.
Some quadratic equations can be factored. For instance,
15x2 + 14x – 8 =0
(3x + 4)(5x – 2) = 0
3x + 4 = 0, or 5x – 2 =0
x = -4/3, 2/5
Unfactorable equations require the quadratic formula.
Quadratic formula: x 
 b  b 2  4ac
2a
The quantity under the radical, b2 - 4ac, is called the discriminant (D).
Assuming a, b, c to be real numbers:
If D > 0, the two roots of the equation are distinct real numbers.
If D = 0, the equation has a real, double root (the roots are equal and real).
If D < 0, the roots of the equation are two complex conjugates (imaginaries).
If r and s are roots of a quadratic equation, r + s = -b/a, and rs = c/a.
If an imaginary number is a solution to a quadratic equation with real coefficients, its conjugate
must be the other solution.
As a side note, the solutions of (x – a)(x – b)(x – c)(x – d)(x – e)(x – f)… = 0 are a, b, c, d, e, f,
…. e.g., If (x + 4) is a factor of a polynomial, -4 is a root. As another example, an equation with
solutions of i, -i, and 3, and leading coefficient –5, is –5(x – i)(x + i)(x – 3) = 0.
2. Complex Numbers
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Remember this, you’ll go far: i =
 1 . Hence, i2 = -1.
A complex number C is a number which can be written in the form a + bi, where a and b are real
numbers and i = 1 . a is the real part of C, and bi is the imaginary part of C.
An imaginary number is a complex number whose imaginary part is not zero (i.e., b  0 ). A
pure imaginary number is an imaginary number whose real part is zero (i.e., a = 0 and b  0 ).
Two complex numbers a + bi and c + di (a, b, c, d are real numbers) are equal if and only if a =
c and b = d.
4 is a real number.
5 + 3i is an imaginary number.
-8i is an imaginary number, as well as a pure imaginary number.
bi is a pure imaginary number as long as b  0 . If b = 0, bi = 0i = 0, which is real.
Remember: each of these numbers is either real or imaginary, but
they are all complex!
3. The Powers of i; Operations on Complex Numbers; Restrictions on the Laws of
Exponents
i1 = i
i2 = -1
i3 = -i
i4 = 1
i5 = i
…
The powers of i are cyclic; they repeat every four powers. i raised to any positive multiple of
four is 1.
i2567 = (i2564) i3 = i3 = -i
(-i)1998 = (-1)1998 i1998 = i1998 = (i1996) i2 = -1
Because any power of i can be expressed in terms of i to the first power or in terms of 1, your
answer should never have any power of i other than i to the first power (i.e., without exponent),
6
and i should never appear in a denominator. Therefore, to rationalize a fraction like
(and you
i
i
must), multiply it by . (The answer is –6i.)
i
Note also that i and –i are reciprocals.
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Addition: (a + bi) + (c + di) = (a + c) + (b + d)i
Subtraction: (a + bi) – (c + di) = (a – c) + (b – d)i
As you can see, adding and subtracting complex numbers is just like adding and subtracting
binomials. So is multiplication. There is a rule, [ (a + bi)(c + di) = (ac – bd) + (ad + bc)i ], but
you would do better just to multiply them as two binomials.
(2 + 2i)(3 – i) = 6 + 6i – 2i – 2i2 = 6 + 6i – 2i – 2(-1) = 8 + 4i
Division entails rationalizing the denominator, to eliminate i from it. Do this by multiplying the
fraction by the conjugate of the denominator over itself. (The conjugate of a + bi is a – bi).
2  2i (2  2i)(3  i) 6  8i  2i 2 4  8i 4  8i 2  4i 2 4





  i
3i
(3  i)(3  i)
32  i 2
9 1
10
5
5 5
Before we move to a new topic, one thing must be said about multiplying and dividing the square
roots of negative numbers. Normally, when you see an expression like 2 3 13 , you can just
multiply the roots together:
2 3 13  78
Whenever you see the square root of a negative number, you must first factor an i out of the
radical, to eliminate the negative sign. Then you can use the rules.
- 2 - 3 - 13  i 2 i 3 i 13  i 78 (which does not equal
 78 !)
Likewise,
5
5  i 5  i 15
, which does not equal



3
3 i 3
3

5
.
3
4. Higher-degree Equations and the Rational Root Theorem; Synthetic Division
First, a word about complex solutions to higher-degree equations. If a + bi is a solution to an
equation with real coefficients, a – bi must also be a solution. This was said earlier, with
respect to quadratic equations, but it holds for any equations of any degree.
There is no simple formula to find the solutions to cubic (third-degree), fourth-degree, etc.,
equations. There is a way, however, to determine whether it has any rational solutions.
The rational root theorem:
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If P(x) is a polynomial whose coefficients (and constant) are integers, C is the set of all factors,
positive and negative, of the constant term of P(x), and D is the set of all factors of the leading
coefficient of P(x), then all rational solutions to P(x) = 0 must be of the form p/q, where p is a
member of C and q is a member of D. (It’s simpler than it sounds.)
Once you have found a root of P(x), you can use synthetic division to divide the factor out of
P(x) and get a new polynomial. Repeat the process until the equation contains only linear and
quadratic factors; you can then use the quadratic formula to find the remaining roots, which
might or might not be rational.
I won’t describe the process fully here, because it usually comes up only on the team round, but I
will demonstrate an example of synthetic division.
Divide x – 2 into 2x3 + 3x2 – 5x – 20.
2
2
2
3
4
7
-5
14
9
-20
18
-2
When dividing by x – a, the divisor in the corner is a. Write the coefficients in order, inserting a
zero in place of any missing terms. Multiply the divisor (2) by the leading coefficient and place
it in the first space under the line. Multiply this number by the divisor, place it under the next
coefficient, add the next coefficient to it, place this number below the line, and repeat until there
are no more coefficients. The last number, -2, is the remainder.
2x 3  3x 2  5x  20
2
 2x 2  7x  9 
x2
x2
If the remainder is zero, that means x – a was a factor of the polynomial, and therefore that a was
one of its roots.
Find all complex roots of x3 – 4x – 15 = 0.
According to the rational root theorem, if there is a rational root, it must be one of 1, 3, 5, –1, –3,
or –5. As it turns out, 3 works:
3
1
1
0
3
3
-4
9
5
-15
15
0
x 3  4x  15
 x 2  3x  5
x3
x3 – 4x – 15 = (x – 3)(x2 + 3x + 5)
x – 3 = 0  x = 3, or
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x2 + 3x + 5 = 0
x
 3  9  20
3
11
 
i
2
2
2
Final answer:
x  3,  3  11 i,  3  11 i
2 2
2 2
If you want more information on solving higher-degree equations, the rational root theorem, and
synthetic division, check out http://mathforum.org/dr.math/problems/drummond8.27.98.html .
Have fun!
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