x 2 5x + 6 Definition 1.2 The domain of an algebraic expression is the set of values that the variable is allowed to take.

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1 .4 Rational Epressions Definition. A quotient of two algebraic epressions is called a fractional epression. A rational epression is a fractional epression where both the numerator and denominator are polynomials. For eample, is a fractional epression while The Domain of an Algebraic Epression is a rational epression. Definition. The domain of an algebraic epression is the set of values that the variable is allowed to take. Eamples:. The epression has domain { 0}, that is, the set of all numbers ecept 0. This is because you cannot divide by 0.. The epression has domain { }, that is, the set of all non-negative numbers. This is because you cannot take the square root of a negative number. 3. The epression has domain { > 0}, that is, the set of all numbers ecept 0. This is because you cannot take the square root of a negative number nor can you divide by Find the domain of + 3. solution: Here is allowed to take any value, so the domain is all real numbers, R. 5. Find the domain of ( )( 3) solution: Here, is allowed to take any value ecept and 3 since both of these would make the bottom zero. Hence the domain is all real numbers ecept and 3 or, written in set notation, {, 3}. 6. Find the domain of ( 5) solution: The square root makes it so we can t let be any negative number, and the 5 in the bottom makes it so that cannot be 5, so in set notation our domain is { 5 and }.

2 Simplifying Rational Epressions Just like when dealing with regular fractions, you can cancel a common term from the numerator and denominator as long as it is not zero: Eample: A C B C if C 0 + ( )( + ) ( )( + ) + + Multiplying and Dividing Rational Epressions if Multiplying and dividing rational epressions is done in an almost identical way to multiplying and dividing ordinary fractions. Multiplication: A B C D AC BD Eample: Division: A B C D A B D C Eample: Adding and Subtracting Rational Epressions ( )( + 3) ( + 4) ) ( + 3) ( + ) ( ) ( + 4)( + ) + 3 ( )( + ) To add or subtract rational epressions, they need to have the same denominator. If they do not, you must first put them under the same denominator before adding or subtracting them: Eamples:. A B + C D A B D D + C D B B AD + BC BD ( + ) ( )( + ) + ( ) ( + )( ) ( )( + ) ( + )( )

3 . 3. Simplify the epression solution: 4. Simplify the epression solution: ( ) y y. + y y a+h a h a+h a h +y y y. a (a+h) a(a+h) h + y y y a a h a(a + h) h ( + y) y( y) a(a + h) 5. Simplify the epression ( + ) / ( + ) / + solution: We should first factor out ( + ) / from the top, as it is the lowest power of that factor present. ( + ) / ( + ) / ( + ) / (( + ) ) ( + ) / ( + ) ( + ) +/ ( + ) 3/ Rationalizing the Denominator In order to rationalize a function with the denominator in the form A + B C we multiply the numerator and denominator by its conjugate A B C. Eample: + + ( )( + ) ( ) As a final note, here are some errors that commonly befall students. Be careful!

4 . (a + b) a + b. a + b a + b 3. a + b a + b 4. a + b a + b 5. a+b a b 6. a + b (a + b).5 Equations and Inequalities Definition.3 An equation is a statement that two mathematical epressions are equal. Solving an equation is to find all values for all the unknowns that make the equation true. An inequality is a statement that two mathematical epressions are <, >,, or each other. Solving an inequality is to find all values for the unknowns that make the inequality true. Eamples: The unknown is and has one solution,.. Solve the equation solution: Definition.4 A linear equation in one variable is an equation equivalent to one of the form a + b 0 where a and b are real numbers and is the variable. Eample: is linear while + 8 is not. If a 0 solving a linear equation is either impossible (if b 0) or trivial (0 0). If a 0, then the solution of a + b 0 is b a.. Solve the equation solution: We isolate the variable by moving all the terms involving to one side and everything else to the other side.

5 . Solve the inequality solution: We can solve inequalities in much the same way that we solve equalities: So the set of that satisfy the inequality is all. In set notation, the solution is { } and in interval notation the solution is (, ]. Note: we solve inequalities in much the same way as equations with one key difference - multiplying each side of an inequality by the same negative number reverses the direction of the inequality. Eample: Solve the inequality 3 < solution: 3 < < 4 ( 6 < 4 ) ( ( 6) > 4 ) 6 6 > 3 Again, the set of all solutions can be written as { > } or (, ). 3 3 Rules of Inequalities. A B A + C B + C.. A B A C B C.

6 3. If C is positive and A B, then AC BC and vice-versa. 4. If C is negative and A B, then AC BC and vice-versa. 5. If A and B are both positive and A B, then A B and vice-versa. 6. If A B and C D, then A + C B + D. Definition.5 A quadratic equation is an equation of the form a + b + c 0 where a, b and c are real numbers with a 0. Solutions of this type of equation are called the zeroes or roots of the polynomial a + b + c. In cases where the quadratic can be factored, finding the solutions is relatively easy. For eample, if we wanted to solve We would notice that ( + 3)( + ) and hence our equation is ( + 3)( + ) 0 Since we have two terms multiplied together to give 0, we know that one of the terms must be zero. Hence, either or + 0 or, in other words, or 3. Thus the roots of are and 3. If the quadratic cannot easily be factored, then to find the solutions you should use the quadratic equation: if a + b + c 0, then the solutions are b ± b 4ac. a In this case, we have that a + b + c must factor as follows: ( ( )) ( ( )) b + a b 4ac b b 4ac + b + c a a Eample: Solve solution: a 3, b 5 and c, so we have ( 5) ± ( 5) 4(3)( ) (3) 5 ± ± 37 6 note that in this case there are two solutions, one from the plus and one from the minus. The quantity b 4ac that appears under the square root is called the discriminant and is denoted. If b 4ac > 0 then the equation has two distinct real solutions. If b 4ac 0 then the equation has eactly one real solution.

7 If b 4ac < 0 then the equation has no real solutions. Eample: Solve + 0. solution: The discriminant is so we have two solutions. They are b 4ac () 4()( ) 8 > 0 b ± a Since 8 4 4, we have Hence we also have that ± ± 8. ± + ( ( + ))( ( )) ( + )( + + ) Non-Linear Inequalities Sometimes you are asked to solve an inequality that is not linear. For eample, suppose you are asked to find the s for which > 0. The techniques mentioned earlier will not work, so we need something else. For our eample, we know that factors, so that our inequality is ( + )( + 3) > 0 So we want to find the values of that make the product of (+) and (+3) positive. We know that for the product of two terms to be postitive, they must either both be positive or both be negative. With that in mind, we set up a table detailing where each of the factors is positive and negative: (, 3) 3 ( 3, ) (, ) ( + )( + 3) The above is what is known as a sign table. The + line, for eample, is read as on (, 3), at 3 and on ( 3, ) ( + is negative while on (, ) it is postitive, and so on. Thus to get the signs of the product function at the bottom of the table we multiply the pluses and minuses down the column. If there are no zeroes and an even number of signs, the result is +. If there are an odd number, the result is. From our table, we can see that ( + )( + 3) is positive on (, 3) and on (, ). Thus the solution set for ( + )( + 3) > 0 is (, 3) (, )

8 In general, to solve non-linear inequalities, follow these steps: Move all the terms to one side. Rewrite the inequality so that all the terms are on one side. If there are multiple quotients, bring them over a common denominator. Factor. Factor the non-zero side as much as you can. Find where each factor is zero. Then split up your table at those values. Draw your sign table. Use test values in the intervals to fill in the signs. Solve. Use the table to determine where the inequality holds. Note that it is very important to move all the terms to one side. Sign tables don t work if you don t do this first. Eamples:. Find the values of such that +. solution: We begin by moving everything to one side ( ) So we have to split up our line at the points 0 and. (, 0) (0, ) (, ) So this is postitive on (0, ). Since we are solving, we need to include any endpoints that make 0. The only endpoint that does that is 0. Thus our solution is S [0, ). (a) Solve

9 (b) Solve solution: (a) We start by getting the left over a common denominator Thus 3 or. 3( + ) + 5 ( + ) 3( + ) + 5 ( + ) ( 3)( + ) (b) Here we have to be a bit more careful. We have to get everything on one side before doing anything ( + ) + 5 ( + ) 3( + ) + 5 (( + )) ( + ) ( + ) ( 3) ( + ) ( 3)( + )) ( + ) Thus we have 4 factors to consider, and must split our table up at the points,, 0, and 3. (, ) (, ) (, 0) (0, 3) (3, ) ( 3)(+)) (+) + + +

10 is positive on (, ), (, 0) and (3, ). It is also 0 at the points 3 and. Hence our inequality holds on So ( 3)(+)) (+) Absolute Value Inequalities S (, ) [, 0) [3, ) Sometimes we are asked to solve an inequality involving an absolute value. Usually this just reduces to solving two regular inequalities. Inequality Equivalent Form Interval < c c < < c ( c, c) c c c [ c, c] > c < c or c < (, c) (c, ) c c or c (, c] [c, ) Eamples: Solve the inequalities.. 5 < solution: From the chart we see this is equivalent to < 5 <. We can solve this by adding 5 to each of the terms: < 5 < 3 < < 7 Thus the set of that satisfies 5 < is the interval (3, 7). In general, if you are given an inequality of the form d < c, then the solution set is an interval centered around c with radius d solution: From the chart we see this is equivalent to or or 3 or 3 corresponds to the interval (, ] while corresponds to the interval [ 3, ). Hence the solution set is (, ] [, ). 3 3