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

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

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

Page 1: 1.5 Quadratic Equations Start p 145 graph and model for #131 & discuss.

1.5 Quadratic Equations

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

Page 2: 1.5 Quadratic Equations Start p 145 graph and model for #131 & discuss.

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.

Page 3: 1.5 Quadratic Equations Start p 145 graph and model for #131 & discuss.

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?

Page 4: 1.5 Quadratic Equations Start p 145 graph and model for #131 & discuss.

Solving a Quadratic Equation by Factoring

1. If necessary, rewrite the equation in the form ax2 bx c 0, moving all terms to one side, thereby obtaining ______ on the other side.

2. Factor.3. Set each factor = zero. (Apply the

zeroproduct principle.)4. Solve the equations in step 3.5. Check the solutions in the __________

equation.

Page 5: 1.5 Quadratic Equations Start p 145 graph and model for #131 & discuss.

Text Example

• Solve 2x2 7x 4 by factoring and then using the zeroproduct principle. (Do not look at notes, no need to write.)

Step 1 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 Step 2 Factor.

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Solution cont.

• Solve 2x2 7x 4 by factoring and then using the zeroproduct principle.

Steps 3 and 4 Set each factor equal to zero and solve each resulting equation.

2 x 1 or x 4 2 x 1 x 4x = 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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Ex: Solve for x:(2x + -3)(2x + 1) = 5 Why can’t we set each factor =5?

ALWAYS begin by

factoring out the GCF.

Q: In the above example, it is not necessary to set the factor 4 = 0, but what if the GCF had been 4x?

Simplify

Set = 0

Factor

Apply zero product principle

Check.

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The Square Root Method

d

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: d

d

25(4 1) 12x

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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 # 54.2 6 5 0x x

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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

a

acbbx

2

42

Given a quadratic equation in the form:

02 cbxaxWe can solve for x by “plugging in” a, b and c:

Derived by completing the square, if interested, see p121.

a>0,

a,b,c integers

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2 5 8x x 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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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 solutionsb2 – 4ac > 0

Graph of y = ax2 + bx + c

Kinds of solutions to ax2 + bx + c = 0

Discriminantb2 – 4ac

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

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Which approach do we use to solve a quadratic equation?

1. Recognize that you have a quadratic equation.2. 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.)

3. 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.)

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The Pythagorean TheoremThe 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)