MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 1 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Chabot Mathematics §5.7 PolyNomial Eqns & Apps

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 2 Bruce Mayer, PE Chabot College Mathematics Review §  Any QUESTIONS About §5.6 → Factoring Strategies  Any QUESTIONS About HomeWork §5.6 → HW MTH 55

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 3 Bruce Mayer, PE Chabot College Mathematics §5.7 Solving PolyNomial Eqns  The Principle of Zero Products  Factoring to Solve Equations  Algebraic-Graphical Connection

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 4 Bruce Mayer, PE Chabot College Mathematics Quadratic Equations  Second degree equations such as 9t 2 – 4 = 0 and x 2 + 6x + 9 = 0 are called quadratic equations  A quadratic equation is an equation equivalent to one of the form ax 2 + bx + c = 0 where a, b, and c are constants, with a ≠ 0

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 5 Bruce Mayer, PE Chabot College Mathematics Principle of Zero Products  An equation AB = 0 is true if and only if A = 0 or B = 0, or both = 0.  That is, a product is 0 if and only if at LEAST ONE factor in the multiplication-chain is 0 i.e.; Need a Zero-FACTOR to create a Zero-PRODUCT

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 6 Bruce Mayer, PE Chabot College Mathematics Example  Solve (x + 4)(x − 3) = 0  In order for a product to be 0, at least one factor must be 0. Therefore, either x + 4 = 0 or x − 3 = 0  Solving each equation: x + 4 = 0 or x − 3 = 0 x = −4 or x = 3  Both −4 and 3 should be checked in the original equation.

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 7 Bruce Mayer, PE Chabot College Mathematics Check for (x + 4)(x − 3) = 0 For x = −4:For x = 3: (x + 4)(x − 3) = 0 (−4 + 4)(−4 − 3) (3 + 4)(3 − 3) 0(−7) 7(0) 0 = 0 0 = 0 True True The solutions are −4 and 3.

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 8 Bruce Mayer, PE Chabot College Mathematics Solve  4(3x + 1)(x − 4) = 0  Since the factor 4 is constant, the only way for 4(3x + 1)(x − 4) to be 0 is for one of the other factors to be 0. That is, 3x + 1 = 0 or x − 4 = 0 3x = −1 or x = 4 x = −1/3  So the solutions to the Equation are x = −1/3 and x = 4 {−1/3, 4}

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 9 Bruce Mayer, PE Chabot College Mathematics Check 4(3x + 1)(x – 4) = 0  For −1 /3:4(3x + 1)(x − 4) = 0 4(3[−1/3] + 1)([−1/3] − 4) = 0 4(−1 + 1)(−1/3 − 12/3) = 0 4(0)( − 13/3 ) = 0 0 = 0  For 4: 4(3x + 1)(x − 4) = 0 4((3(4) + 1)(4 − 4) = 0 4(13)(0) = 0 0 = 0  The solutions are − 1/3 and 4

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 10 Bruce Mayer, PE Chabot College Mathematics Solve  3y(y − 7) = 0  SOLUTION 3  y  (y − 7) = 0 y = 0 or y − 7 = 0 y = 0 or y = 7  The solutions are 0 and 7 The Check is Left to the Student –Should be easily “EyeBalled”

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 11 Bruce Mayer, PE Chabot College Mathematics Factoring to Solve Equations  By factoring and using the principle of zero products, we can now solve a variety of quadratic equations.  Example: Solve x 2 + 9x + 14 = 0  SOLUTION: This equation requires us to FIRST factor the polynomial since there are no like terms to combine and there is a squared term. THEN we use the principle of zero products

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 12 Bruce Mayer, PE Chabot College Mathematics Solve  x 2 + 9x + 14 = 0  Factor the Left Hand Side (LHS) by Educated Guessing (FOIL Factoring): x 2 + 9x + 14 = 0 (x + 7)(x + 2) = 0 x + 7 = 0 or x + 2 = 0 x = −7 or x = −2.  The Tentative Solutions are −7 and −2 Let’s Check

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 13 Bruce Mayer, PE Chabot College Mathematics Check  x = −7& x = −2 For −7:For −2: x 2 + 9x + 14 = 0 x 2 + 9x + 14 = 0 (−7) 2 + 9(−7) + 14 (–2) 2 + 9(–2) − − − − = 0 0 = 0 True True  Thus −7 and −2 are VERIFIED as Solutions to x 2 + 9x + 14 = 0

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 14 Bruce Mayer, PE Chabot College Mathematics Example  Solve x 2 + 9x = 0  SOLUTION: Although there is no constant term, because of the x 2 -term, the equation is still quadratic. Try factoring: x 2 + 9x = 0 → see GCF = x x(x + 9) = 0 x = 0 or x + 9 = 0 x = 0 or x = − 9  The solutions are 0 and − 9.

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 15 Bruce Mayer, PE Chabot College Mathematics Caveat Mathematicus  CAUTION  CAUTION: We must have 0 on one side of the equation before the principle of zero products can be used.  Get all nonzero terms on one side of the equation and 0 on the other  Example: Solve: x 2 − 12x = − 36

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 16 Bruce Mayer, PE Chabot College Mathematics Solve  x 2 − 12x = −36  SOLUTION: We first add 36 to BOTH Sides to get 0 on one side: x 2 − 12x = − 36 x 2 − 12x + 36 = − = 0 (x − 6)(x − 6) = 0 x − 6 = 0 or x − 6 = 0 x = 6 or x = 6  There is only one solution, 6.

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 17 Bruce Mayer, PE Chabot College Mathematics Standard Form  To solve a quadratic equation using the principle of zero products, we first write it in standard form: with 0 on one side of the equation and the leading coefficient POSITIVE.  We then factor and determine when each factor is 0.

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 18 Bruce Mayer, PE Chabot College Mathematics Example  Functional Eval  Given f(x) = x x Find a such that f(a) = 1.  SOLUTION Set f (a) = 1 f (a) = a a + 26 = 1 a a + 25 = 0 (a + 5)(a + 5) = 0 a + 5 = 0 so a = −5 The check is left for you & I to do Later

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 19 Bruce Mayer, PE Chabot College Mathematics Example  Solve 9x 2 = 49  SOLUTION:9x 2 = 49 9x 2 − 49 = 0 ► Diff of Sqs: (3x) 2 & 7 2 (3x − 7)(3x + 7) = 0 3x − 7 = 0 or 3x + 7 = 0 3x = 7 or 3x = − 7 x = 7/3 or x = − 7/3  The solutions are 7/3 & − 7/3

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 20 Bruce Mayer, PE Chabot College Mathematics Solve: 14x 2 + 9x + 2 = 10x + 6  SOLUTION: Be careful with an equation like this! Since we need 0 on one side, we subtract 10x and 6 from Both Sides to get the RHS = 0 14x 2 + 9x + 2 = 10x x 2 + 9x − 10x + 2 − 6 = 0 14x 2 − x − 4 = 0 (7x − 4)(2x + 1) = 0 7x − 4 = 0 or 2x + 1 = 0 7x = 4 or 2x = −1 x = 4/7 or x = −1/2  The solutions are 4/7 and −1/2.

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 21 Bruce Mayer, PE Chabot College Mathematics Solve Eqns by Zero Products 1.Write an equivalent equation with 0 on one side, using the addition principle. 2.Factor the nonzero side of the equation 3.Set each factor that is not a constant equal to 0 4.Solve the resulting equations

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 22 Bruce Mayer, PE Chabot College Mathematics Parabola intercepts  Find the x-intercepts for the graph of the equation shown. y = x 2 + 2x  8  The x-intercepts occur where the plot crosses y = 0 Parabola  Thus at the x-intercepts y = 0 = x 2 + 2x − 8  So Can use the principle of zero products

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 23 Bruce Mayer, PE Chabot College Mathematics Parabola intercepts cont.  SOLUTION: To find the intercepts, let y = 0 and solve for x. y = x 2 + 2x  8 0 = x 2 + 2x − 8 0 = (x + 4)(x − 2) x + 4 = 0 or x − 2 = 0 x = −4 or x = 2  The x-intercepts are (−4, 0) and (2, 0). Parabola

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 24 Bruce Mayer, PE Chabot College Mathematics −x 2 − x + 6 = 0  Solve with Graph  Solve  Recall from Graphing that the x-axis is the Location where y = 0  Thus on Graph find x for y = 0  Solns: x = −3 and x = 2

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 25 Bruce Mayer, PE Chabot College Mathematics Example  Find x-Intercepts  Find the x- intercepts for graph (at Left) of Equation  At intercepts y = 0; so Use ZERO Products:

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 26 Bruce Mayer, PE Chabot College Mathematics Example  Find x-Intercepts  At Intercepts y = 0, so y = 0 = 2x 2 + 3x − 9  FOIL-factor the Quadratic expression 0 = 2x 2 + 3x − 9 = (2x − 3)(x + 3) = 0  By ZERO PRODUCTS (2x − 3) = 0 or (x + 3) = 0  Solving for x (the intercept values): x = 3/2 or x = −3

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 27 Bruce Mayer, PE Chabot College Mathematics Example  x-Intercepts

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 28 Bruce Mayer, PE Chabot College Mathematics PolyNomial Fcns and Graphs  Consider, for example the eqn, x 2 − 2x = 8. One way to begin to solve this equation is to graph the function f (x) = x 2 − 2x. Then look for any x-value that is paired with 8, as shown at Right x y y = 8

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 29 Bruce Mayer, PE Chabot College Mathematics PolyNomial Fcns and Graphs  Equivalently, we could graph the function given by g(x) = x 2 − 2x − 8 and look for the values of x for which g(x) = 0. These values are what we call the roots, or zeros, of a polynomial function Root-1Root-2

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 30 Bruce Mayer, PE Chabot College Mathematics Problem Solving  Some problems can be translated to quadratic equations, which we can now solve.  The problem-solving process is the same as for other kinds of problems.

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 31 Bruce Mayer, PE Chabot College Mathematics The Pythagorean Theorem  Recall Pythagorus’ Great Discovery:  In any right triangle, if a and b are the lengths of the legs and c is the length of the hypotenuse, then a 2 + b 2 = c 2 a b c

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 32 Bruce Mayer, PE Chabot College Mathematics Example  Screen Diagonal  A computer screen has the dimensions (in inches) shown below.  Find the length of the diagonal of the screen.

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 33 Bruce Mayer, PE Chabot College Mathematics Example  Screen Diagonal  Familiarize. A right triangle is formed using the diagonal and sides of the screen. x + 6 is the hypotenuse with x and x + 3 as legs.  Translate. Applying the Pythagorean Theorem, Use the Diagram to translate as follows: x 2 + (x + 3) 2 = (x + 6) 2 a 2 + b 2 = c 2

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 34 Bruce Mayer, PE Chabot College Mathematics Example  Screen Diagonal  Carry out. Solve the equation by: (x + 3)(x – 9) = 0 x + 3 = 0 or x – 9 = 0 x = –3 or x = 9 2x 2 + 6x + 9 = x x + 36 x 2 – 6x – 27 = 0 x 2 + (x 2 + 6x + 9) = x x + 36

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 35 Bruce Mayer, PE Chabot College Mathematics Example  Screen Diagonal  Check. The integer −3 cannot be the length of a side because it is negative. For x = 9, we have x + 3 = 12 and x + 6 = 15. Since = 225, the lengths determine a right triangle. Thus 9, 12,and 15 check.  State. The length of the diagonal of the screen is 15 inches

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 36 Bruce Mayer, PE Chabot College Mathematics Example  Area Allotment  A LandScape Architect designs a flower bed of uniform width around a small reflecting pool. The pool is 6ft by 10ft. The plans call for 36 ft 2 of plant coverage. How WIDE should the Border be?  Familiarize with Diagram

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 37 Bruce Mayer, PE Chabot College Mathematics Example  Area Allotment  Now LET x ≡ the Border Width  Translate using Diagram  The OverAll Width is 6ft + 2x Length is 10ft + 2x

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 38 Bruce Mayer, PE Chabot College Mathematics Example  Area Allotment

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 39 Bruce Mayer, PE Chabot College Mathematics Example  Area Allotment  Carry Out Zero Products  Since a Length can Never be Negative, Discard −9 as solution

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 40 Bruce Mayer, PE Chabot College Mathematics Example  Area Allotment  State: The reflecting pool border width should be 1 ft 12 ft 8 ft  Check: 2·(12ft·1ft) + 2·(6ft·1ft) = 24ft ft 2 = 36ft 2 

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 41 Bruce Mayer, PE Chabot College Mathematics Example  Rocket Ballistics  A Model Rocket is fired UpWards from the Ground. The Height, h, in feet of the rocket can be found from this equation:  Find the time that it takes for the Rocket to reach a height of 48 feet –Where t is time in seconds

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 42 Bruce Mayer, PE Chabot College Mathematics Example  Rocket Ballistics  Familiarize: We must find t such that h(t) = 48. Thus Substitute 48 for h(t) in the ballistics equation  Carry Out: Subtract 48 from both sides

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 43 Bruce Mayer, PE Chabot College Mathematics Example  Rocket Ballistics Divide Both Sides by −16 Write in Standard form Use QUADRATIC Formula with a = 1, b = −5, and c = 3 Prime  Not Factorable

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 44 Bruce Mayer, PE Chabot College Mathematics Example  Rocket Ballistics Approximate the Sq-Roots  Since “What goes UP must come DOWN” The Rocket will reach 48ft TWICE; Once while blasting-UP and again while Free-Falling DOWN

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 45 Bruce Mayer, PE Chabot College Mathematics Example  Rocket Ballistics  State: The rocket will climb to 48ft in about 0.7 seconds, continue its climb, and then it will descend (fall) to a height of 48ft after a total flite time of 4.3 seconds as it continues its FreeFall to the Ground

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 46 Bruce Mayer, PE Chabot College Mathematics WhiteBoard Work  Problems From §5.7 Exercise Set 74, 80, 82  Model Rockets

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 47 Bruce Mayer, PE Chabot College Mathematics All Done for Today Blast Off!

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 48 Bruce Mayer, PE Chabot College Mathematics Bruce Mayer, PE Licensed Electrical & Mechanical Engineer Chabot Mathematics Appendix –

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 49 Bruce Mayer, PE Chabot College Mathematics Graph y = |x|  Make T-table

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 50 Bruce Mayer, PE Chabot College Mathematics

MTH55_Lec-26_sec_5-7_PolyNom_Eqns-n-Apps.ppt 51 Bruce Mayer, PE Chabot College Mathematics Quadratic Formula