SAT Math THE FINAL MILE Copyright Β© 2021 Harsh Manghnani and Nathan Quadras All rights reserved. No parts of this book may be reproduced without explicit written permission from the authors. SAT is a trademark registered and owned by the College Board, which is not affiliated with, and does not endorse, this product. Contact [email protected] for requests or details. Special thanks to our mentors, Mr. Satya Prakash Singh and Mr Dilip Biswas, for their unparalleled guidance throughout the entire process of formulating this book. Their help was truly invaluable to us. Contents 1. Quadratic Equations.......................................................................................................1 2. Systems of Equations.....................................................................................................4 3. Inequalities.....................................................................................................................7 4. Exponents and Radicals................................................................................................11 5. Linear and Exponential Functions................................................................................13 6. Percentage...................................................................................................................17 7. Ratio and Proportion....................................................................................................18 8. Rates.............................................................................................................................19 9. Straight Lines.................................................................................................................20 10. Functions......................................................................................................................23 11. Absolute Value Functions.............................................................................................28 12. Complex Numbers........................................................................................................30 13. Expressions...................................................................................................................33 14. Solving Equations.........................................................................................................35 15. Angles...........................................................................................................................36 16. Trigonometry................................................................................................................38 17. Triangles.......................................................................................................................42 18. Circles...........................................................................................................................47 19. Volume.........................................................................................................................49 20. Statistics.......................................................................................................................51 21. Probability....................................................................................................................54 22. Linear Models...............................................................................................................55 Quadratic Equations Different Forms of Quadratic Functions: Standard form: π(π₯) = ππ₯ 2 + ππ₯ + π a: It controls the degree of curvature of the graph and whether the graph opens upwards or downwards. If a< 0 , then the graph opens downwards, but if a> 0, then the graph opens upwards. Larger the value of a, narrower the parabola. Conversely, smaller the value of a, wider the parabola. b: It controls the location of the axis of symmetry of the parabola. Along with the value of b, it also depends on the coefficient a. Letβs have a look at the following situation: 1.) a is positive and b is negative, it shifts to the right. 2.) a is negative and b is positive it shifts to the right. 3.) a is positive and b is positive it shifts to the left. 4.) a is negative and b is positive it shifts to the left. c: This represents the yintercept of the parabola. Factored form: π(π₯) = π(π₯ β π1 )(π₯ β π2 ) π1 and π2: These represent the roots of the parabola. π1 +π2 : gives the x coordinate of the vertex. In other words, the midpoint of the 2 roots represents the x coordinate of the vertex. π1 +π2 π( 2 ): This represents the y coordinate of the parabola. π(0) : This represent the yintercept. Vertex form: π(π₯) = π(π₯ β β)2 + π h: x coordinate of the vertex. This can also be viewed as the coordinate for the vertical line of symmetry k: y coordinate of the vertex. This can also be seen as the maximum or minimum value depending on the value of a. if a< 0 then k represents the maximum value of the parabola while if a> 0, k represents the minimum value of the parabola 1 Quadratic Formula: βπΒ±βπ2 β4ππ The quadratic formula, which is , allows us to find the roots of quadratic equations. 2π βπ+βπ2 β4ππ βπββπ 2 β4ππ One root is represented by and the second root is represented by 2π 2π Relation between roots and coefficients: π βππππππππππ‘ ππ π₯ Sum of roots: For the equation π(π₯) = ππ₯ 2 + ππ₯ + π = 0 , the sum of the roots is β = π ππππππππππ‘ ππ π₯ 2 π ππππ π‘πππ‘ π‘πππ Product of roots: For the equation π(π₯) = ππ₯ 2 + ππ₯ + π = 0 , the product of the roots is = π ππππππππππ‘ ππ π₯ 2 Therefore, a quadratic equation can be expressed in the following form π₯ 2 β (π π’π ππ ππππ‘π )π₯ + πππππ’ππ‘ ππ ππππ‘π =0 π π π₯ 2 β (β π) + π=0 π₯ 2 β ππ₯ + π=0 Discriminant: The number of roots of a parabola depends on the value of the discriminant. The discriminant of a quadratic in the form of π(π₯) = ππ₯ 2 + ππ₯ + π is π₯ = π 2 β 4ππ. Letβs look at the following situations: Discriminant Value Roots Graph It has 2 real and distinct π₯>0 roots/solutions π₯<0 It has no real roots/Solutions 2 π₯=0 It has one real solution 3 Systems of Equations Equation with No Solutions A system of equation is said to have no solution if there is no common solution for both equations. This is the case because these lines are parallel lines and will thus never intersect. How to Identify a system of equation that has No Solutions? Systems with No Solutions have these characteristics. 1. The Lines have the same slope 2. They have different y intercepts In such a situation convert both the equations into standard form. The ratio of the coefficients should ππ ππ ππ be as follows: = β . Letβs look at an example. ππ ππ ππ How many Solutions does the equation below have? 3x + 4y = 6 6x + 8y = 9 A. One B. Two C. No solution D. Infinitely many solutions 3π₯ β 4π¦ β 6 = 0 6π₯ + 8π¦ β 9 = 0 3 4 = 6 8 1 1 = 2 2 In the case of the y intercept: 6 1 1 = β 9 3 2 System with infinitely many Solutions A system of equations is said to have infinitely many solutions if the 2 equations are equivalent to each other and are represented by the same line when plotted on a graph. 4 How to identify a system of equations with infinitely many solutions: 1) They have the same slope 2) They have the same yintercept π1 π1 π1 In other words, = = . π2 π2 π2 Look at this example: How many solutions does this system of equations have? 2x + 3y = 5 4x + 6y =10 2 3 5 = = π¦ 6 10 1 1 1 = = 2 2 2 These 2 equations represent the same line and thus have infinitely many solutions. Equations with one Solution These are the equation of graphs which intersect at one point and thus have only one solution. How to Identify a system of equations with one solution 1. They have different slopes 2. They have different y intercepts. π1 π1 π1 β β π2 π2 π3 Letβs look at an example: How many solutions does this system of equations have? π¦ = 2π₯ + 2 π¦ = β3π₯ + 7 On arranging the equation into standard form, we get: 2π₯ β π¦ + 2 3π₯ β π¦ β 7 Now on arranging the coefficients and constants into their respective ratios, we get: 2 2 β 1β 3 7 Therefore, the above system of equations has only one solution. 5 Number of Equations Expression Graph π1 π1 π1 No Solutions = β π2 π2 π2 Infinite number of π1 π1 π1 = = solutions π2 π2 π2 π1 π1 π1 β β One solution π2 π2 π3 6 Inequalities Inequalities resemble normal equations, except for the fact that when you multiply or divide by a negative sign, you need to reverse the inequality. β2π₯ < 2 βπ₯ < 1 π₯ > β1 As you can see, we first divide both sides by 2, and then to get the range of the values of x, we multiply both sides by 1, reversing the sign in the process. Try doing this question and see if you get the correct answer: Question: ππ + π β€ ππ + ππ Ans: π₯ β₯ β8 Note: Only reverse the sign when multiplying or dividing both sides by a negative number, NOT if a negative number simply exists in the inequality. If 2π₯ < β2, π₯ < β1, not π₯ > β1. If a question asks you to pick which of the given options is a solution to the inequality, try to plot it on a number line and find out. For example, if the inequality you get is π₯ > β5, and the given options are A. 7 B. 6 C. 5 D. 4 You can try plotting it on a number line, and youβll see that the only number lying in the shaded region is 4. 5 is not a solution to the given inequality as it is a βstrictly greater than inequalityβ and has an open circle at 5 indicating that it is not inclusive. Thus, the answer is D. 4 Another type of problem quite common in the SAT is when a set of information is given, and a student must pick the right inequality. For example, it might be given that a man burns 5 calories for every mile he bikes and burns 10 calories for every mile he runs, and that he must burn at least 100 calories a day. In this case, if the miles he bikes in a day is given by p and the miles he runs in a day is given by q, then the inequality is 5π + 10π β₯ 100 < Less than > Greater than β€ Less than or equal to β₯ Greater than or equal to = Equal to 7 Note: If a<b: (i) a+c<b+c (ii) ac<bc (iii) ka<kb, for k>0 (iv) ka>kb, for k<0 yβ€ π y< 5 Here, the gray region represents the region Here, the gray region represents the region where the y value is less than or equal to 5. where the y value is less than 5. The dotted line indicates that coordinates where y=5 are not solutions to the inequality. xβ€5 Here, the gray region represents the region where the x value is less than or equal to 5. 8 2x3y> 6 Here, the gray region represents the region where the value of 2x3y is greater than 6. The dotted line indicates that values of x and y for which 2x3y=6 are not solutions to the inequality. 2x3yβ₯ 6 Here, the gray region represents the region where the value of 2x3y is greater than or equal to 6. 9 x+yβ€ π xyβ₯ 7 Here, there are two inequalities on the graph: x+yβ€6 and xyβ₯7. The light gray regions represent the regions where either x+yβ€6 or xyβ₯7, and the dark gray region represents the intersection of both the inequalities, i.e., where x+yβ€6 and xyβ₯7. 10 Exponents and Radicals 1. Any number to the power 1 is the number itself, and any number to the power 0 is 1. a. π₯ 1 = π₯ b. π₯ 0 = 1 2. π₯ π β π₯ π = π₯ π+π π₯π 3. = π₯ πβπ π₯π 4. (π₯ π )π = (π₯ π )π = π₯ ππ 5. (π₯π¦)π = π₯ π β π¦ π π₯ π π₯π 6. (π¦) = π¦π 1 7. π₯ βπ = π₯ π 8. βπ₯ Γ βπ¦ = βπ₯π¦ 9. If you have a negative sign INSIDE the parentheses, you must raise it to the power of the parentheses as well, i.e., (βπ₯)π is written as (β1 β π₯)π = (β1)π β π₯ π 10. If the bases are equal, the powers can be equated, i.e., If π₯ π = π₯ π , π = π, π₯ β 1 1 1 1 π 3 π π Note: A square root, i.e., βπ₯, can be written as π₯ 2 . Similarly, βπ₯ = π₯ 3 , βπ₯ = π₯ π and βπ₯ π = π₯ π . To simplify a radical, find its prime factors, pair them up, and take them out of the root. For instance, β192 = β2 Γ 2 Γ 2 Γ 2 Γ 2 Γ 2 Γ 3 =2 Γ 2 Γ 2 Γ β3 = 8β3 Example: If π₯ ππ₯ β π₯ ππ₯ = π₯ 45 and π + π = 15, we can find the value of x by using the laws of exponents. π₯ ππ₯ β π₯ ππ₯ = π₯ 45 π₯ ππ₯+ππ₯ = π₯ 45 (π + π)π₯ = 45 (Equating the exponents as bases are equal) 15π₯ = 45 π₯=3 11 Note: On the SAT, we always take the positive value of a root, i.e., β4 is only equal to 2, not 2. However, if x2=4, then x= +2 or 2. π₯ 2 = 64 has two solutions, +8 and 8. However, β64 has just one solution, +8. 12 Linear and Exponential Functions What is Linear Growth? When the independent variable undergoes a change in constant increments and the dependent variable grows each time by a constant amount, the growth is referred to as linear growth. In other words, for an equal change in the intervals of in the x value, the y value changes by a constant amount. Linear growth can be expressed through the means of a linear function. π(π₯) = ππ₯ + π Let us determine what each component of a linear equation means: m: This represents the slope of the linear function. It shows the change in the value of y for a change in the value of x. In other words, it is the constant rate of change in the value of y. b: This represents the y coordinate of the y intercept. In other words, it is the value of the linear function when the value of x is 0. It shows the initial value of the function, therefore when the question asks you to give the starting value of the function, find the y intercept of the function. Note: If you see c instead of b, donβt worry! Both b and c represent the y intercept of the function. Linear Growth Linear Decay 13 How do you find the equation for a linear function when given a table? π π(π) 1 7 3 11 4 13 First, we must determine the slope of the linear function which can be done by simply choosing any two pairs from the table. consider points (1,7) and (3,11). On calculating the slope, we get: π¦2 β π¦1 11 β 7 4 π= = = =2 π₯2 β π₯1 3β1 2 So far, our equation is π¦ = 2π₯ + π . In order to determine the y intercept of the linear function, we must insert any point into the equation. Consider point (1,7) which when inserted into the function, 7=2+π gives us: 5= π Now that we have the y intercept value, we can form our equation which is: π¦ = 2π₯ + 2 How to find the Linear equation from a graph. To solve this question, we must make use of the slope intercept formula. On looking at the graph we can see that the line l passes through the point (0,5) on the y axis. Therefore, for line l, b=5. To calculate the slope, we must select to points on the graph through which line l passes through. Take points (6,8) and (8,9). Using these points, the slope of the graph comes to 9β8 1 1 π = 8β6 = 2 . Therefore, the equation of the line l is π¦ = 2 π₯ + 5 14 What is Exponential Growth? When the independent variable undergoes a change in constant increments and the dependent variable grows each time by the same rate. In other words, the y value increase at a rate relative to the previous value. π (π₯) = π(π)π₯ Exponential Growth Exponential Decay Let us analyze each part of the exponential function: a: This represents the initial value of the function. In other words, it is the y intercept of the exponential function. The point is usually expressed as π(0). In order to determine the value of a, you have to simply look at the point on the graph where the function intersects the y axis or where the x value is 0 in the case of tables. b: This represents the growth/decay factor of the exponential function. If the value of b is 0 < π < 1 then the function is experiencing exponential decay, but if the value of b is π < 1, then the function is experiencing exponential growth. In order to calculate the value of b, the growth factor, you must first find the percent growth or decay rate, which is expressed as a decimal. For example, if the function increases by 20% for every increase in the x value, then b is written as follows: π = (1 Β± π) π = (1 + 0.20) π = 1.20 However, if the function decreased by 20% for every increase in the x value, the function experiences exponential decay and is expressed as π = (1 Β± π) follows: π = (1 β 0.20) π = 0 β 80 15 Finding an exponential function from a table x π(π) 1 c 2 d 3 e π β π= = π π On finding the value of b and inserting it into the expression, you have to simply insert a given point into the expression to find the value of a. 16 Percentage π₯ A percent is exactly as it sounds, βper centβ, i.e., per 100. Therefore, π₯% = 100. π₯ π₯% ππ π¦ = Γπ¦ 100 Remember: x% of y= y% of x The SAT might ask you to extrapolate data from a smaller dataset and apply it to a bigger set of data. In that case, simply find the percentage for the smaller dataset and multiply it with the total number of entries in the bigger dataset. For example, if in a class of 10, there are 4 girls and 6 boys, the SAT may ask you to find the number of girls in a school of 500 students. In that case, we find the percentage of girls in the given class, i.e., 40%, and multiply it with the total number of students to get 40% of 500 or 200 girls. In case of a percentage increase or decrease, the new value can be found by multiplying the original π₯ π₯ value by (1 + 100) or (1 β 100) respectively. In case of a multiple percentage change in price, the value is changed according to the new price. For example, if the price of a $10 pack of chocolates is increased 10 10 11 9 by 10% then decreased by 10%, the new price is 10 β (1 + ) β (1 β ) = 10 β β = $9.9, not $10 100 100 10 10 as you would think. Note: A 20% increase in the value of x can be written as (1+0.2)β x= 1.2x and similarly, a 20% decrease in the value of x can be written as (10.2)β x= 0.8x. πππ€ ππππ’πβπππ ππππ’π Percentage Increase= πππ ππππ’π πππ€ ππππ’πβπππ ππππ’π Percentage Decrease= πππ ππππ’π As you see, both the percentage increase and decrease have the original value in the denominator, NOT the new value. Simple Interest PrincipalβRateβTime πβ π β π Simple Interest= 100 = 100 Amount= Principal+ Simple Interest Compound Interest π ππ‘π(ππ π πππππππ) Amount= Principal (1+ )nt , where n is the number of times interest is compounded π per unit βtβ and t is the number of years. Compound Interest= AmountPrincipal Note: Annually: n=1 Semiannually: n=2 Quarterly: n=4 17 Ratio and Proportion Let area of a rectangle be length x breadth, i.e., π΄ =πΓπ If the length is tripled, we get π΄πππ€ = 3π Γ π = 3(π Γ π) =3π΄ This is known as proportion. We see that by tripling the length, the area is tripled, or the new area is 300% of the new area. Now letβs take another example. If we take a square of side a, area of the square is π΄ = π2 And if the side length is doubled, π΄πππ€ = (2π)2 = 4π2 = 4π΄ In this case, the area is quadrupled, i.e., it becomes 22 times the original area or the area is 400% the original area. Thus, a proportion is a ratio that is used to determine the change in the value of an equation when π₯ one of its components is altered. If a value, say x, is halved, you substitute 2 in the original equation to find the change in the value of the equation as a percent. If multiple components are changed, substitute all the changes to find the net change in the value of the equation. If π = 3π3 Γ π 4 and the value of P is increased by 30%, the value of Q is decreased by 70%, then the net change in the value of M is ππππ€ = 3(1.3π)3 Γ (0.3π)2 = 3 Γ 2.197π3 Γ 0.09π 2 = 0.59319 Γ π3 Γ π 2 = 0.59319 Γ π Thus, the new value of M is 59.319% of its original value, or the value has decreased by 40.681%. 18 Rates Rate is equal to the work done or the number of items produced (whatever the question may be) per unit time. For example: (i) if a man eats 5 cakes in 10 hours, we say that he eats 5/10= 0.5 cakes an hour. In other words, his rate of eating cake is 0.5 cakes per hour. (ii) If an artist paints 5 paintings in 20 days, we say that he paints 5/20= 0.25 paintings a day. In other words, his rate of painting paintings is 0.25 paintings per day. (iii) If a student studies 1 topic in 5 hours, we say that he studies 1/5= 0.20 topics an hour. In other words, his rate of studying topics is 0.20 topics per hour. Using the same first example, a rate can be used to determine the cakes eaten for any number of hours, or alternatively the time taken to eat a certain number of cakes. The same can be done with the remaining two examples as well! For example, if we need to find the number of hours taken to eat 500 cakes, we find it as follows: ππ’ππππ ππ πππππ πππ‘ππ Rate= ππππ π‘ππππ(ππ βππ’ππ ) ππ’ππππ ππ πππππ πππ‘ππ Time Taken (in hours) = π ππ‘π 500 = = 1000 hours 0.5 Similarly, if we need to find the number of cakes a man can eat in 20 hours, we find it as follows: ππ’ππππ ππ πππππ πππ‘ππ Rate= ππππ π‘ππππ(ππ βππ’ππ ) Number of cakes eaten= Rate Γ Time taken (in hours) = 0.5 Γ 20 = 10 cakes. Thus, rates can be used alongside any other SAT concept to ask a question as it is a very versatile concept that must be thoroughly understood. 19 Straight Lines 1.) Slope of a straight line: When given 2 distinct points  (x1 y1) and (x2 y2)  on a line the formula is as follows: π¦2 βπ¦1 π₯π¦ π ππ π = = = π‘ππ π π₯2 βπ₯1 π₯π₯ π π’π When a line passes through the origin then you must simply insert the y value of one point over the x value of that point. A slope is considered positive if it goes up towards the right and negative if it goes down towards the right. Positive Slope Negative Slope 2.) Slope intercept form: y = mx + c m= slope of the line c= yintercept 3.) Pointslope form y β y1 = m (x x1) m=slope of the line 4.) Standard form of an equation Ax + By + C = 0 In most straightline questions on the SAT, the options will appear in this form. Note: The slope of a horizontal line is 0, and the slope of a vertical line is undefined. 20 Expressing a straight line as a function A function is an equation for which only be one specific yvalue is produced for a given xvalue. In other words, all the points on the straight lines in the Cartesian plane can be expressed as (π₯, π(π₯)) Consider the equation π¦ = 2π₯ + 3 At the point π₯ = 2, the value of y is π¦ = 2(2) + 3 π¦ =4+3 π¦=7 Therefore, that point can be expressed as (2,7) Remember: f (0) is simply the y intercept. Lines with equal slopes are called parallel lines, and a line whose slope is the negative 1 reciprocal of another line is said to perpendicular to that line. That is, π1 = β π 2 Type of Line Expression Graph y=2 y=k where k is the Horizontal Line units above or below the origin on the y axis x= k x=2 where k is the Vertical Line units of the line left or right to the origin. 21 m1 = m2 Parallel lines Perpendicular 1 π1 = β Lines π2 22 Functions β’ If f(x)= x2, then f(a)= a2, where a is any real number or variable. β’ If f(x)= x2 and g(x)= x3, then f(g(x))= f(x3)=x6 β’ In the SAT, if youβre given the value of f(x) and asked to find the value of a sum of the function several times with different values of x, simply substitute the given value into the equation for f(x). Example: If f(x)= x2, then f(0)+f(1)+f(2)= 0+1+4= 5 β’ If a function is given in the form of a fraction and you need to find the values where the function is undefined, equate the denominator to zero. The values you obtain when you solve for x are the xvalues where the function is undefined. β’ Note: Always work your way inside out when given a function in a function, i.e., if youβre asked to find the value of f(g(h(i(x)))), then you first find the value of i(x), take the resulting value as the domain(xvalue) for h(x), find that value, take it as the domain of g(x), and so on. β’ f(x) is just another way to say y. β’ If a function is given and it is said to cross a point, say (x,y), then you can substitute the x and y values into the given function to answer the question as it may be. β’ The roots of the equation are simply the xvalues when y=0, or the xvalues where the function crosses the xaxis. The yintercept, denoted by c, is the value of y when x=0. β’ If you are given a function and asked to choose which graph could be a graph of that function, find a point where you can substitute the x and y values into the function and if the equation holds, that is the correct option. Example: If f(x)= 2x2+2, you might have 4 graph options. If you plug in x=0, you get y=2. Thus, look at the graph and whichever graph passes through (0,2) is the right graph. If more than one graph holds true, try choosing another point as well to figure out which option is correct. β’ A oneone function is a function in which every y value has exactly one x value mapped to it. β’ A manyone function is a function in which at least one y value has more than one x value mapped to it. 23 Vertical Line Test The vertical line test tests whether a given expression is a function or not. By the definition of a function, each xvalue must correspond to only one specific yvalue. However, multiple xvalues can have the same yvalue as long as a specific xvalue only corresponds to one yvalue. In the graph of f(x)=x, each x value corresponds to exactly one yvalue. Try it out yourself, no xvalue will give multiple yvalues! In this case, the graph of x2+y2=9 is not a function as for a specific xvalue represented by blue (x=2), there are two different corresponding yvalues (2.236 and 2.236). 24 Transformation of Functions Transformation Method Graph Shifting up or down A graph can be shifted up or down by adding or subtracting a constant at the end of the function. For example, f(x)+3 will shift the graph of f(x) up by 3 units, f(x) being x2 f(x)+3 f(x) Shifting left or right A graph can be shifted left or right by adding or subtracting a constant inside the parentheses of the function. For example, f(x+2) will shift the graph of f(x) left by 2 units, f(x) being x2 f(x+2) f(x) 25 Reflecting across the xaxis If f(x)= x2, then f(x) will be or yaxis the graph of f(x) reflected across the xaxis. f(x) f(x) Similarly, if f(x)= x+2, then f(=x) will be the graph of f(x) reflected across the y axis. f(x) f(x) Absolute values As an absolute function is a function that gives the absolute value, regardless f(x)= x of the sign, any part of the graph that dips below the xaxis will be reflected back above. f(x)= x 26 Even and Odd Functions An even function is one in which if the variable, say x, is replaced by its negative, in this case x, the function remains the same. Example: Let f(x)= x2, then f(x)= (x)2= x2. As the function remains the same, it is an even function. An odd function is one in which if the variable, say x, is replaced by its negative, in this case x, the function becomes the negative of the original function. Example: Let f(x)= x3, then f(x)= (x)3= x3. As the function is the negative of the original function, it is an odd function. A function is neither even nor odd (NENO) if f(x) is neither equal to f(x) nor equal to f(x). Example: If f(x)= x3x2, then f(x)= (x)3(x)2= x3x2. As it is not equal to f(x) or f(x), it is neither even nor odd. Even Odd f(x)=f(x) f(x)= f(x) Symmetric about yaxis Symmetric about origin 27 Absolute Value Functions The absolute value is, simply, the distance of the value from zero on the number line. The absolute value of 1 is 1, and of 1 is also 1. In other words, if the given function has a value less than 0, then the absolute value of that function is the function multiplied by 1. It is denoted by modulus (π₯). Quite often, this concept comes clubbed with inequalities on the SAT. Example: If π₯ < 4, find the values of x that satisfy the given inequality. As π₯ is less than 4, x must be between 4 and 4. Thus, we can write that β4 < π₯ < 4 Example: If π₯ + 1 < 4, find the values of x that satisfy the given inequality. β4 < π₯ + 1 < 4 As we want to find the values of x, we subtract 1 from each part of the inequality to get x at the center. β5 < π₯ < 3 Note: For π₯ β₯ 0, π₯ = π₯ and for x < 0, π₯ = βπ₯ Another important thing to note in absolute values is the graph of modulus of x. If y= π₯, we know that even for negative values of x, the value of y is positive. Thus, the graph of π₯ is as follows: As you can see, the value of y is never negative, as the absolute value of x is always positive. Now, if the function is y= π₯ + 1, then we know that y will be zero at x=1. Thus, 28 The graph has been shifted by 1 unit to the left. In general, if y= π₯ + π, then the graph is shifted k units to the left and if y= π₯ β π, the graph is shifted k units to the right. 29 Complex Numbers A number expressed as π + β π, where a is the real part and if is the imaginary part, is called a complex number. If π§ = β π then the number is called a purely imaginary number while if π§ = π then the number is called a purely real number. π + β π Imaginary Real Part Part Note: If a is any positive real number, then ββπ = βπβ If βπ β βπ = βππ , then at least one of them is positive If βπ β βπ = ββππ , then π < 0 and π < 0 Powers of β β = ββ1 β 2 = β1 β 3 = ββ β 4 = 1 β 4π = 1 β 4π+1 = β β 4π+2 = β1 β 4π+3 = ββ β 4π+4 = 1 How to find the value of a complex number raised to a nonnegative integer First divide the power of the complex number by 4, depending on the remainder obtained your answer will be one of the following: Value of Remainder Value of complex number 0 1 1 β 2 1 3 β 30 Addition of 2 Complex Numbers Expand the expression and combine the like terms. If π§1 = 4 + 5β and π§2 = 7 + 8β ,then π§1 + π§2 is =(4 + 5β ) + (7 + 8β ) =11 + 13β Subtraction of Complex Numbers Expand the expressions and combine the like terms. If π§1 = 6 + 8β and π§2 = 2 + 4β , then π§1 β π§2 is =(6 + 8β ) β (2 + 4β ) =6 + 8β β 2 β 4β =4 β 4β Multiplication of Complex Numbers Perform FOIL on the expression If π§1 = 3 + 4β and π§2 = 7 + 8β then π§1 π§2 is =(3 + 4β )(7 + 8β ) =21 + 24β + 28 + 32β 2 =32β 2 + 24β + 49 Division of Complex Numbers Multiply the numerator and denominator by the conjugate of the denominator. π§1 If π§1 = 2 + 7β and π§2 = 2 + 3β then π§2 is (2+7π) (2β3π) = (2+3π) β (2β3π) 4β6π+14β21π 2 = 4β9π 2 4+14+21β6π = 4+9 39β6π = 13 6 =3 β β 13 31 Conjugate The conjugate of a complex number is when you inverse the sign of the imaginary part, i.e., if the original expression is 2+3i, then its conjugate will be 23i. The conjugate can be used when rationalizing in questions where roots are involved. π§ = π + β π π§Μ = π β β π 32 Expressions Combining Like Terms When trying to simplify a given expression, you must take note of the exponents of each term, as that is what determines whether it can combine with another term or not. For instance, π₯(π₯ 4 β 3π₯ 3 + π₯ β 5) β 4π₯ 2 π₯ 5 β 3π₯ 4 + π₯ 2 β 5π₯ β 4π₯ 2 π₯ 5 β 3π₯ 4 β 3π₯ 2 β 5π₯ As you can see, the term 4x2 is subtracted from x2 as they have the same exponent, that is, 2. Thus, to combine like terms, the powers or exponents must be the same. FOIL Another concept to keep in mind is multiplying an expression in the form of (a+b)(c+d). It is done by using FOIL, which stands for First, Outer, Inner, and Last. (π + π)(π + π) = ππ + ππ + ππ + ππ Some common special expansions are: (a+b)2= a2+b2+2ab (ab)2= a2+b22ab (a+b)(ab)= a2b2 Note: A common mistake that students make is when given an expression like 4x2 9y2, they factorize it to give (4x+9y)(4x9y) which is incorrect as they forget to convert 4x2 to (2x)2 Dividing Fractions When dividing by a fraction, you can multiply the numerator by the reciprocal of the denominator to get the same answer, i.e., π π = π β π = ππ π π π ππ π Splitting Fractions 33 If youβre given an expression containing the sum of various values in the numerator, and an expression in the denominator, you can split the fraction by dividing each segment of the numerator by the denominator, that is, π+π π π = + π π π Note: The same is not the case with denominators, that is, the sum of various values in the denominator cannot be split into segments with the same numerator. This holds true only for such an expression in the numerator. If a question mentions an undefined value, remember that an expression is undefined if the denominator is equal to 0, or if there is a negative value under a root. 34 Solving Equations Though this concept is relatively easier and doesnβt have much content to cover, it is nonetheless very important as it is a basic concept used for many other questions. We will not be discussing the elimination method and substitution method as they are quite trivial and learnt in early high school. As discussed earlier, combining like terms is essential in order to solve equations, as more often than not, all except one variable cancels out, giving you, its value. Similarly, you must manipulate the equation by take squares, square roots, etc. in order to simplify the expression and solve the given question. Note: You must always take the principal value (positive value) of a root, that is, β4 = 2, NOT 2 unless specified in the question. Factorizing is an important concept to keep in mind that is used virtually everywhere on the SAT. It can be done by either taking out common factors, noticing special expansions, or by splitting the middle term. If you are unable to factorize the expression, you can use the quadratic formula as a last resort. βπ Β± βπ 2 β 4ππ π₯= 2π Cross multiplying is another basic concept that you must keep in mind, and is quite selfexplanatory. π π If π = π, then ad=bc Example 1: If 16π₯ + 8π¦ + 2π§ = 16, find 8x+4y+z. This is one of the simplest problems on the SAT, yet one that confuses many students. One thing to note is that unless the question asks for the value of a specific variable, you must not restrict yourself to solving for one specific variable, you should first try solving for all of them as a unit. What this means is that in the given question, you can easily find 8x+4y+z by taking 2 common from the given expression. Then, by crossmultiplying, we get that 8x+4y+z= 8. 2(8π₯ + 4π¦ + π§) = 16 8π₯ + 4π¦ + π§ = 8 Matching coefficients Another important concept in the SAT is matching coefficients. What this means is that if Ax2+Bx+C= 2 Dx +Ex+F, then A=D, B=E, and C=F. Example 1: If 2π₯ 2 + 5π₯ + 3 = π΄π₯ 2 + π΅π₯ + πΆ, find A, B and C. You see that the coefficient of x2 on the lefthand side is 2, and on the righthand side is A. Equating the coefficients, we get that A=2. Similarly, B=5 and C=3. Note: If a question like (x3)(x+6)(x9) = ax3+bx2+cx+d is given and asks you to find the value of d, donβt multiply the entire expression! Just multiply the last terms, i.e., 3*6*9, to get 162, the value of d. 35 Angles Exterior Angle Theorem This theorem states that the exterior angle in a triangle is equal to the sum of the two interior angles opposite to it. In this example, dΒ°= aΒ°+bΒ°, i.e., the exterior angle is equal to the sum of the two interior opposite angles. Angles in Parallel Lines 36
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