Merced College; Don Power   

 

 

2.1  First Degree Equations and Applications    

 

Basic procedure for solving 1st degree equations in 1 variable:

            [Collect like terms whenever possible]

Clear grouping symbols

            General technique:  Distributive law

            Occasionally, you can divide by the coefficient on a term with parentheses

                        Ex:  2(x+3) = 6

            Clear fractions (optional, but recommended)

                        General technique:  Multiply every term by the LCD of all the fractions

                        For proportions, you can cross-multiply

            Move terms: (+ or -) to separate terms

                        Terms with the variable go on one side of the equals sign, and

                                    terms without the variable go to the other side of the equals sign

            Factor (if the variable appears in more than one term)

            Divide by the coefficient of the variable

                        (or multiply by the reciprocal, if you didn't clear fractions)

 

   Ex:  2(3/x + 1) +6 (3/(4x) - 2) = 1  Like X14, but with variables in the denominators.

 

Same steps when solving for a letter:

            x/2 + a/3 = -b(c-dx)      Solve this for x

 

Applied probs (Hwk):   See examples, and see step-by-step setups given in text for X49-52

            X51, 82:  Given perimeter and relationship between length and width...

X49, 54:  Score on the next test needed to get a certain average.

X57, 58:  Markdown, where we are to find the old price.

X61, 62:  Mixed investments:  See Ex 9

X66:  Weighted avg:  .3(1st)+.3(2nd)+.4(Final)=Course average.  Weights must add to 1.

X69:  Mixture, drain & replace:  See Ex 10

X74:  Distance, Rate, Time:  See Ex 12

X77:  Work:  See Ex 13   (think of the rate:  jobs per hour)

X81:  Filling a tub that leaks:  Think of the rates:   tubs/minute in and out.

 

2.1A    Variation

 

Steps:

            1.  Set up the basic variation equation.

                        Use "k" as a constant of variation (constant of proportionality)

                        k goes into the numerator

                  Direct or joint variation:  Factors go into the numerator, along with the k.

                  Inverse variation:  Factors go into the denominator

            2.  Substitute a matched set of values and solve for k

            3.  Rewrite the variation equation from step 1, but replace the letter k by the value from step 2

            4.  Use the equation from step 3 to solve application problems.

 

Ex for class:  #41:  F = k*W*S2 / R.  Force, weight of vehicle, speed of vehicle, radius of curve

            Known case:  {F,W,S,R} = {1500kg,1000kg,50kph,200m}

            Case to be solved:  Same car, radius 320m at 100kph.

 

2.2  Quadratic Equations and Applications  

 

Ex 1,2  Solve by factoring

            "Zero product" property:  If a product is 0, then at least one of the factors has to be 0.

            Technique:

                        Set all = 0

                        Factor

                        Set each factor = 0

                        Solve the resulting smaller equations

            Ex:  Solve 5x2 = 2x

 

Ex 3     This example, (z−2)2=5, uses the "Square Root Property:"  If A2 = B, then A = ħsqrt(B)

 

Ordinary completing the square:

            x2 − 8x + 5 = 0     The coefficient of x2 must be 1

            Move the constant:  x2 − 8x = − 5

            When we complete the square, we add a constant (b/2)2 to both sides to make the left side a perfect square:  x2 − 8x + ___ = − 5 + __

            We will work to get the equation into this form:  ( ___ ħ ___ )2 = ___    

                        The first term inside the parenthesis has to be x.

                        The second term inside the parentheses is b/2, which we already calculated.

 

Text uses completing the square to derive the quadratic formula

 

Ex 4-7 Calculating with the quadratic formula

 

            Simplifying the radical, then the fraction:  2x2 − 6x − 5 = 0

 

            "No real solution"   x2 − x + 3 = 0     Later we'll write two complex solutions

 

Ex 8:    Use of calculator to approximate the solutions.

            X45-48

 

Find number and type of solutions with the discriminant b2 − 4ac

           

Ex 9-11  Applications

            #76  Typical box problem

            #87-90  Height problems.  See background in Example 7:  h = -16t2 - v0t + h0

                        v0 is the initial upward velocity (so it is negative if the object is thrown downward)

                        h0 is the initial height

 

Ex 12:  Rational equations that result in quadratics (when you clear the fractions):  #31-34

 

 #55-59  Solving for a letter:  apply the quadratic formula, where the a, b, and c may be variable expressions

 

 

2.3  Solving Eqns Graphically and Numerically   

Solving graphically:  FACT:  the real solutions of a 1-var eqn in the form expr(x) = 0

are the x-intercepts of the graph of the 2-var eqn of the form y = expr(x)  [why? Let y=0...]

Ex:  Solve sqrt(x^2-5) = abs(x-3) + ln x     Note: set =0; adjust window (or trace to zero)

Caution:  roots may exist that are not shown on the screen

Contrast with intersection method:  weakness:  intersec may be above or below the window.

Equation solver:  weakness: it only finds one root (nearest to a guess you type in - so graph first)

            86:  2nd SOLVER; for the eqn, give y1=0, and type guess, leave cursor on x and <SOLVE>

            83:  [catalog] SOLVE(y1,x,guess)

or MATH <Solver> For eqn enter 0=y1, x=guess, green ENTER

Text mentions polynomial solver (86, not 83) -- finds complex roots -- see Techno Tip in text

"USE YOUR BRAIN":  Ex 6 & 7:  cases where a little prelim alg makes the calcr work possible/easy

            Ex 6: sqrt(x^4+x^2-2x-1)=0  Implies radicand=0, or:  Clear the radical first

            Ex 7:  fraction = 0:  This implies the num = 0  (denom is irrelevant unless it's also 0)

 

 

2.3  Other Equations and Applications   

Quadratic type -- by subs, to make it quadratic (e.g. some 4th degree eqns)

            Ex:  6x^4-7x^2=3 has 2 real, 2 imaginary solutions/roots.

Graphical sol will work if other sols are not possible -- impact on test?

Ex 6:  Crossing river at an angle:  See #51

Radical eqns

            Ex:  #27:  sqrt(2x-5) = 1 + sqrt(x-3)

Abs value eqns Ex 8 is like the nasty ex I gave you earlier.

 

2.4  Polynomial, Radical, and Absolute Value Equations

 

 

 

2.5  Linear Inequalities   

Basic:  Do #8:  5-3x>7x-3 

Calculate

Graph sol for > and for ³.   Result is a single interval, infinite on one end

Ineq vs interval notation.

Compound Ineq:  Do #20:  2x+5£4-3x<1-4x

Calculate

Graph sol.  Result is a single interval. Open, closed, or half-open?

Ineq vs interval notation.

Basic Abs value inequalities     (note:  won't work if you have a variable outside the abs value)

            < or £  Do #36  abs(5x-1)<3

Calculate

Graph sol.  Result is a single interval. Open, closed, or half-open?

Ineq vs interval notation.

            > or ³  Two intervals:  Modify #36:  abs(5x-1)>3

Calculate

Graph sol.  Result:  two intervals. Open, closed, or half-open?

Ineq vs interval notation.

            Graphical interpretation:  the distance from 5x to the center (at +1) is < or > than 3.

Alternate approach

            1.  Solve the related equation (find critical points)

            2.  Determine whether the inequality is true or false at each critical point (mark solid or open dot)

            3.  Partition the x-axis into intervals based on those critical points -- use interval notation

            4.  Select the intervals that make the inequality true.  How:

                        Select a test point inside each interval.

                        Test the test point in the inequality.

 

2.6  Polynomial and Rational Inequalities    Use the alternate approach described for 2.4:  Do #48 (modified):  1/(x-1) £ -1/(x+2)

1.  Solve the eqn:  Both polynomials and rationals: set equal to 0, [add fractions], then factor:

                        Critical points are those that make the numerator or denominator 0

            2. Determine whether the inequality is true or false at each critical point (mark solid or open dot)

                        If denominator = 0, the ineq is false at the Critical point

                        If numerator = 0, the ineq may be true (for £ or ³) or false at critical point (for < or >)

            3.  Partition the x-axis into intervals based on those critical points -- use interval notation

            4.  Select the intervals that make the inequality true.  How:

                        Select a test point inside each interval.

                        Test the test point in the inequality.

           

Graphical solution -- relate to the "Alt approach"  Same Example:  Gr y=1/(x-1) +1/(x+2)

              Root is at -.5, VA at -2 and 1.

            Put 0 on the right side of the inequality before graphing

            For critical points:

Find the zeros/roots of y = ____;

                        Also find points where func is undefined (VA)  Easiest:  solve for denom = 0

            Partition x-axis based on the critical points:  use [  ]  (  )  as appropriate.

            Select intervals based on the graph:

                        For > or ³, select intervals on which the graph is above the x-axis

                        For < or £, select intervals on which the graph is below the x-axis

 

 

Return to:  Merced College; Don Power               Updated 08/31/05 by Don Power