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1.5 Quadratic Equations Start p 145 graph and model for #131 & discuss.

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**Definition of a Quadratic Equation**

A quadratic equation in x is an equation that can be written in the standard form ax2 + bx + c = 0 where 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.

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**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?

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**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.

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Text Example Solve 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 - 4 2x2 + 7x - 4 = 0 Step Factor.

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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 = 0 2 x = 1 x = -4 x = 1/2 Steps 5 check your solution (by putting each solution back into the ORIGINAL equation to see if it yields a TRUE statement.

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**(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 by factoring out the GCF. Simplify Set = 0 Factor Apply zero product principle Check. Q: In the above example, it is not necessary to set the factor 4 = 0, but what if the GCF had been 4x?

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The Square Root Method If 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:

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**What term should be added to the binomial **

Text Example What term should be added to the binomial x2 + 8x so that it becomes a perfect square trinomial? Then write and factor the trinomial. x2 + 8x +____2 = (x + ____)2 Note: this is still an expression, not an equation. Do (factor by completing the square- see instructions next slide first) p 144 # 54.

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Completing the Square If 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)2 That 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.

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**Solving by the Quadratic Formula**

Given a quadratic equation in the form: a>0, a,b,c integers We can solve for x by “plugging in” a, b and c: Derived by completing the square, if interested, see p121.

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**Ex: Solve by using the quadratic formula:**

Put into form Identify a,b,c Plug in Simplify Common errors: Not writing the division bar all the way. -b means – (whatever b is!). In this case –(-8) = 8.

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**The Discriminant and the Kinds of Solutions to ax2 + bx +c = 0**

No x-intercepts No real solution; two complex imaginary solutions b2 – 4ac < 0 One x-intercept One real solution (a repeated solution) b2 – 4ac = 0 Two x-intercepts Two unequal real solutions b2 – 4ac > 0 Graph of y = ax2 + bx + c Kinds of solutions to ax2 + bx + c = 0 Discriminant b2 – 4ac

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**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 = 0 a. 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.)

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**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, then a2 + b2 = c2 do # p144:105, 138 (set up)

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