The absolute value (modulus) of a number Given a real number x, its absolute value or modulus is dened as x if x is positive x = 0 if x = 0 x if x is negative For example, 6 = 6, 10 = ( 10) = 10. The absolute value of a number is always positive! Note: We have x 2 = x. Dana Mackey (DIT) MATH 1 / 27
Graphs involving the modulus Sketch the following graphs: y = x y = 2x 1 x = y + 1 Dana Mackey (DIT) MATH 2 / 27
Quadratic equations A quadratic equation is an equation of the form ax 2 + bx + c = 0 where a, b and c are real numbers and a 0. There are many methods for solving a quadratic equation. For example, use the formula x = b + b 2 4ac 2a or x = b b 2 4ac 2a The solutions of a quadratic equation are also called roots. Dana Mackey (DIT) MATH 3 / 27
It appears that every quadratic equation has two roots. In fact, If the quantity inside the square root, b 2 4ac is positive, there are indeed two solutions. For example, x 2 5x + 6 = 0 has solutions x = 3 and x = 2. If the quantiy b 2 4ac is zero, then the solutions are equal so we say the equation has a repeated root. For example, x 2 + 4x + 4 = 0 has the repeated root x = 2 If the quantiy b 2 4ac is negative, the quadratic equation has no real solutions. For example, x 2 + 2x + 2 = 0 has no real roots Dana Mackey (DIT) MATH 4 / 27
s Two resistors are in parallel. The total resistance has been measured at 2Ω, and one of the resistors is known to be 3Ω more than the other. What are the values of the two resistors? Find two numbers whose sum is equal to 3 and product is 4. In general, to nd two numbers with sum S and product P we have to solve the quadratic: x 2 S x + P = 0 Dana Mackey (DIT) MATH 5 / 27
Solving a quadratic by completing the square Recall the formula (x + a) 2 = x 2 + 2ax + a 2 Consider the expression x 2 + 6x + 5. The rst two terms, x 2 + 6x remind us of the expansion (x + 3) 2 = x 2 + 2 3x + 3 2 = x 2 + 6x + 9 Hence, x 2 + 6x + 5 = (x 2 + 6x + 9) 4 = (x + 3) 2 4 To solve x 2 + 6x + 5 = 0 write (x + 3) 2 = 4 so x + 3 = ±2 so x = 1 or x = 5. Dana Mackey (DIT) MATH 6 / 27
Solve the following quadratic equations using square completion x 2 + 4x + 5 = 0 2x 2 5x + 3 = 0 x 2 5x + 2 = 0 9x 2 + 16 = 0 x 2 + 6x + 11 = 0 Note: The formula for the roots of a quadratic was obtained from writing the square completion: ax 2 + bx + c = 0 so ( x + b 2a ) 2 + c b 2 a 4a 2 = 0 Dana Mackey (DIT) MATH 7 / 27
Sum and product of roots Suppose the quadratic equation ax 2 + bx + c = 0 has roots x 1 and x 2. Suppose the sum of these roots is S and their product is P. Then we know that x 1 and x 2 are the roots of the equation x 2 Sx + P = 0 On the other hand, the original quadratic can be written as x 2 + b a x + c a = 0 so comparing the coecients we get the formulas Z Sum of roots =x 1 + x2 = b a Product of roots=x1x2 = c a Dana Mackey (DIT) MATH 8 / 27
Factorization of quadratic equation Every quadratic expression with two real roots x 1 and x 2 can be factorized that is, written as a product of two factors as follows: ax 2 + bx + c = a(x x 1 )(x x 2 ) x 2 6x 7 = (x 7)(x + 1) 2x 2 3x + 1 = 2(x 1)(x 1 ) = (x 1)(2x 1) 2 4x 2 4x + 1 = 4(x 1 2 )2 = (2x 1) 2 Note: If the quadratic has no roots, then it cannot be factorized. Dana Mackey (DIT) MATH 9 / 27
Note: Factorization is a method for solving quadratic equations. Once a quadratic is factorized, the roots are obvious! Solve the following quadratic equations 3x 2 + 4x = 0 x 2 5x 14 = 0 12x 2 20x + 3 = 0 12x 2 23x + 10 = 0 Dana Mackey (DIT) MATH 10 / 27
The following formulae are useful Z x 2 + 2ax + a 2 = (x + a) 2 x 2 2ax + a 2 = (x a) 2 x 2 a 2 = (x a)(x + a) Note: The expression x 2 + a 2 cannot be factorised! Factorise the following quadratic expressions: x 2 + 6x + 9 x 2 25 x 2 + 36 Dana Mackey (DIT) MATH 11 / 27
Solving a linear equation and a quadratic simultaneously To solve these, make x or y the subject of the linear equation and then substitute into the quadratic equation to get another quadratic equation. Note that, two simultaneous equations, one linear and one quadratic have in general two pairs of solutions. Solve the equations x 2y = 7; x 2 + 4y 2 = 37 Dana Mackey (DIT) MATH 12 / 27
Plotting the quadratic expression y = ax 2 + bx + c To plot the graph of y = x 2 we can use a table of values x -3-2 -1 0 1 2 3 y = x 2 9 4 1 0 1 4 9 The graph of the quadratic expression y = ax 2 + bx + c is called a parabola or quadratic curve. Note that this quadratic curve has a vertex at (0,0) which is its minimum point. Dana Mackey (DIT) MATH 13 / 27
Plot the graph of the quadratic y = x 2 + 2x 2 Using a similar table of values we obtain x -3-2 -1 0 1 2 3 y = x 2 + 2x 2-17 -10-5 -2-1 -2-5 Note that this curve has a vertex at (1,-1), which is its maximum point. Dana Mackey (DIT) MATH 14 / 27
In general, to plot a quadratic curve y = ax 2 + bx + c, follow the steps: 1 Check the sign of a: if a > 0 then the curve points up; if a < 0 the curve points down; 2 Determine the intersections with the x and y axes (also called the x and y-intercepts); 3 Find the vertex by one of the following methods: 1 By completing the square. If the quadratic is written as y = a(x m) 2 + n then (m, n) are the coordinates of the vertex. 2 Get the x coordinate from the formula x = b 2a by substitution. and the y coordinate Dana Mackey (DIT) MATH 15 / 27
Plot the graph of the quadratic y = x 2 x 6 Plot the graph of the quadratic y = 4x 2 + 4x 1 Dana Mackey (DIT) MATH 16 / 27
Intersections of a line with a parabola To nd the coordinates of the intersection points between a line and a parabola, we solve two simulataneous equations, one linear and one quadratic. Draw the graphs of y = 6 x and y = x 2 + x 2 and determine the coordinates of their intersection points Dana Mackey (DIT) MATH 17 / 27
Quadratic inequalities For which values of x is x 2 x 2 < 0? First we solve the quadratic equation x 2 x 2 = 0 which can be factorised as (x 2)(x + 1) = 0 so x = 2, x = 1 From the shape of the quadratic curve, we see that the required interval is ( 1,2), that is 1 < x < 2. Dana Mackey (DIT) MATH 18 / 27
For which values of x is x + 1 < x 2 x 2? Show that, for all values of x we have x 2 + 2x + 3 > 0 Dana Mackey (DIT) MATH 19 / 27
Consider the quadratic expression y = ax 2 + bx + c. If a > 0 and the quadratic has two distinct roots then the quadratic is negative for x between the two roots and positive elsewhere. If a > 0 and the quadratic has no roots or a repeated root, then the quadratic is always positive. If a < 0 and the quadratic has two distinct roots then the quadratic is positive for x between the two roots and negative elsewhere. If a < 0 and the quadratic has no roots or a repeated root, then the quadratic is always negative. Dana Mackey (DIT) MATH 20 / 27
Polynomials A polynomial is an expression involving powers of x of the form where a 0, a 1,...a n are real numbers. P(x) = a 0 + a 1 x + a 2 x 2 + a n x n The degree of the polynomial is given by the highest power of X. 5 can be thought of as a polynomial of degree 0 (a constant expression) x 9 is a polynomial of degree 1 (a linear expression) x 2 2x + 1 is a polynomial of degree 2 (a quadratic) x 3 + x 2 + 6x 10 is a polynomial of degree 3 (a cubic) Dana Mackey (DIT) MATH 21 / 27
Roots of polynomials Given a polynomial P(x) = a 0 + a 1 x + a 2 x 2 + a n x n, a number x 0 is called a root of the polynomial if it is a solution of the equation P(x) = a 0 + a 1 x + a 2 x 2 + a n x n = 0 This means that P(x 0 ) = 0 (if we substitute x 0 for x in the polynomial, we get zero). 6 is a root of the polynomial x 6 2 is a root of the polynomial x 2 3x + 2 1 is a root of the polynomial x 3 x 2 + x 1 2 is a root of the polynomial x 8 64 Dana Mackey (DIT) MATH 22 / 27
A polynomial of degree n can have at most n roots. x 4 10x 2 + 9 has 4 roots: ±1 and ±3 x 4 + 1 has no real roots at all x 4 + 8x 2 9 has only two roots: ±1. How do we nd roots for a polynomial of degree higher than 2? There are generally no methods other than guessing. If the coecient of the highest power is 1, a good guessing strategy is to look for roots among the divisors of the constant term. Can you nd any roots for x 4 3x 3 + x 2 10x + 21? Dana Mackey (DIT) MATH 23 / 27
Polynomial factorisation To factorise a polynomial P(x) means to write as P(x) = Q(x) R(x) where Q and R are polynomials of smaller degree than P. x 2 5x + 6 = (x 2)(x 3) x 3 1 = (x 1)(x 2 + x + 1) x 4 3x 3 + x 2 10x + 21 = (x 3)(x 3 + x 7) Note that if a polynomial has a linear factor of the form x x 0, this means that x 0 is a root of the polynomial! Dana Mackey (DIT) MATH 24 / 27
Dividing a polynomial by a linear factor x x0 Suppose we know that x 0 is a root of the polynomial P(x). Then x x 0 is a factor for P(x), that is P(x) = (x x 0 ) Q(x) How do we nd the other factor, Q(x)? This is very similar to number factorisation. If we know that 6 is a factor of 48, to nd the other factor we divide 48 by 6, so 48 = 6 8. Same with polynomials: to nd Q(x), we have to divide P(x) by x x 0 The procedure is called polynomial long division. Divide x 2 7x + 12 by x 3. Divide x 4 3x 3 + x 2 10x + 21 by x 3. Dana Mackey (DIT) MATH 25 / 27
Procedure for factorising polynomials Try to guess a root x 0 (for example, by looking at the divisors of the constant term) Then x x 0 is a factor. Divide the polynomial by x x 0 to get another polynomial Q(x) Repeat the procedure: look for a root of Q(x), then divide Q(x) by the corresponding linear factor, etc. Stop when the polynomial is a product of linear factors or else when you get a factor which cannot be factorised further. Factorise the polynomial x 3 3x 2 10x + 24. Dana Mackey (DIT) MATH 26 / 27
Polynomial long division Note that, when dividing a polynomial of degree n by a linear factor x x 0, the degree of the resulting polynomial is equal to n 1. Long division is also used for dividing a polynomial P(x) by another polynomial Q(x), which has degree less than that of P, but is not necessarily linear. Divide x 4 + 5x 3 2x 2 + 20x 24 by x 2 + 4. It can often happen that a polynomial Q(x) does not divide evenly into a polynomial P(x) and we end up with a remainder. Divide x 4 7x 3 + 6x 39 by x 7. Dana Mackey (DIT) MATH 27 / 27