Presentation on theme: "1.5 Quadratic Equations Start p 145 graph and model for #131 & discuss."— Presentation transcript:
1 1.5 Quadratic EquationsStart p 145 graph and model for #131 & discuss.
2 Definition of a Quadratic Equation A quadratic equation in x is an equation that can be written in the standard formax2 + bx + c = 0where a, b, and c are real numbers with a not equal to 0. A quadratic equation in x is also called a second-degree polynomial equation in x.
3 The Zero-Product Principle If the product of two algebraic expressions is zero, then at least one of the factors is equal to zero.If AB = 0, then A = 0 or B = 0.Q: Will this work for any other number, such as AB=5?
4 Solving a Quadratic Equation by Factoring If necessary, rewrite the equation in the form ax2 + bx + c = 0, moving all terms to one side, thereby obtaining ______ on the other side.Factor.Set each factor = zero. (Apply the zero-product principle.)Solve the equations in step 3.Check the solutions in the __________ equation.
5 Text ExampleSolve 2x2 + 7x = 4 by factoring and then using the zero-product principle. (Do not look at notes, no need to write.)Step Move all terms to one side and obtain zero on the other side. Subtract 4 from both sides and write the equation in standard form.2x2 + 7x - 4 = 4 - 42x2 + 7x - 4 = 0Step Factor.
6 Solution cont.Solve 2x2 + 7x = 4 by factoring and then using the zero-product principle.Steps 3 and Set each factor equal to zero and solve each resulting equation.2 x - 1 = 0 or x + 4 = 02 x = 1 x = -4x = 1/2Steps 5 check your solution (by putting each solution back into the ORIGINAL equation to see if it yields a TRUE statement.
7 (2x + -3)(2x + 1) = 5 Why can’t we set each factor =5? ALWAYS begin by Ex: Solve for x:(2x + -3)(2x + 1) = 5 Why can’t we set each factor =5?ALWAYS begin byfactoring out the GCF.SimplifySet = 0FactorApply zero product principleCheck.Q: In the above example, it is not necessary to set the factor 4 = 0, but what if the GCF had been 4x?
8 The Square Root MethodIf u is an algebraic expression and d is a positive real number, then u2 = d has exactly two solutions.If u2 = d, then u = or u = -Equivalently,If u2 = d then u = We only use this method if the variable is originally contained within a “squared part”. Ex: x2-8=12 or (2x-4)2 –5=20. Can you think of a counter example? Do:
9 What term should be added to the binomial Text ExampleWhat term should be added to the binomialx2 + 8x so that it becomes a perfect square trinomial? Then write and factor the trinomial.x2 + 8x +____2= (x + ____)2Note: this is still an expression, not an equation.Do (factor by completing the square- see instructions next slide first) p 144 # 54.
10 Completing the SquareIf x2 + bx is a binomial, then by adding (b/2) 2, which is the square of half the coefficient of x, a perfect square trinomial will result. That is,x2 + bx + (b/2)2 = (x + b/2)2That is, take half of the coefficient of the x term, square it, and add it to each side.Then take +/- the square root of each side. (Square root method.)Note: this is really just using the fact that the square of a binomial results in a perfect square trinomial. We are just “completing” the perfect square trinomial.
11 Solving by the Quadratic Formula Given a quadratic equation in the form:a>0,a,b,c integersWe can solve for x by “plugging in” a, b and c:Derived by completing the square, if interested, see p121.
12 Ex: Solve by using the quadratic formula: Put into formIdentify a,b,cPlug inSimplifyCommon errors:Not writing the division bar all the way.-b means – (whatever b is!). In this case –(-8) = 8.
13 The Discriminant and the Kinds of Solutions to ax2 + bx +c = 0 No x-interceptsNo real solution;two complex imaginary solutionsb2 – 4ac < 0One x-interceptOne real solution(a repeated solution)b2 – 4ac = 0Two x-interceptsTwo unequal real solutionsb2 – 4ac > 0Graph ofy = ax2 + bx + cKinds of solutionsto ax2 + bx + c = 0Discriminantb2 – 4ac
14 Which approach do we use to solve a quadratic equation? 1. Recognize that you have a quadratic equation.If the variable is isolated within the “squared part”, isolate the squared part, take +/- square root of each side, then isolate the variable. (Square root method.)Otherwise set = 0a. If it is EASY to factor, factor, set each factor equal to zero and solve for the variable (Factoring method.)b. If it is NOT easy to factor, plug a, b, and c into the quadratic formula and simplify (Quadratic formula method.)4. If it says to solve by completing the square, do so (Completing the square method.)
15 The Pythagorean Theorem The sum of the squares of the lengths of the legs of a right triangle equals the square of the length of the hypotenuse.If the legs have lengths a and b, and the hypotenuse has length c, thena2 + b2 = c2 do # p144:105, 138 (set up)