Lecture Note
University
Indiana University South BendCourse
MATH-A 100 | Fundamentals of AlgebraPages
5
Academic year
2020
Rose G
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25
Math A-100 IUSB Sec. 6.6 Detailed Notes – Week 11 Linear vs. Quadratic equations As you recall from Chapter 3, a linear equation is one that makes a straight line on the grid. The basic equation for lines is y = mx + b highest exponent on “x” is 1. A polynomial equation with degree 2 (i.e. has a 2 nd power as the highest exponent on “x”) are called “Quadratic Equations” and have the basic form of y = ax 2 + bx + c. (polynomial equations with higher exponents have other names – “Quadratic” tells us the equation has an exponent of 2). Using factoring to solve equations Recall from Sec 5.3: The Steps for Factoring: 1) Look for a GCF (greatest common factor). Pull it out if one is there. This will make whatever is left easier to work with because the numbers will be smaller. (sec 5.3) 2) Look at the number of terms a) If there are 4 terms, use Grouping (Sec 5.3) b) If there are 3 terms, use the “ac” method (Sec 5.3) c) If there are 2 terms, it is a special case (Sec 5.3) The next logical step is to apply these methods to solving equations, instead of just simplifying polynomial expressions. Recall the difference between expressions and equations equations have the = sign, so you can find out what the value of “x” is. To solve an equation with polynomials: a) Setting the equation equal to 0 b) Factoring the polynomial c) Applying the Zero -Factor principle Example 1: Solve 2x 2 + 11x + 15 = 0 a) Already done for this one b) Factoring 2x 2 + 11x + 15 gives you (2x + 3)(x + 5) = 0 c) the Zero -Factor principle the only way to multiply two values together and get an answer of 0 is if one (or both) of the value(s) IS 0!!! So, set each set of ( ) equal to 0 (2x + 3) = 0 or (x + 5) = 0 Solving gives you x = -3/2 or x = -5 Answer: x = -3/2, -5
Example 2: Solve 7x 2 + 48x = 7 a) 7x 2 + 48x = 7 becomes 7x 2 + 48x – 7 = 0 b) Factoring 7x 2 + 48x – 7 gives you (7x - 1)(x + 7) = 0 c) the Zero -Factor principle set each set of ( ) equal to 0 (7x - 1) = 0 or (x + 7) = 0 Solving gives you x = 1/7 or x = -7 Answer: x = -1/7, -7 NOTE: setting each set of ( ) = 0 only works for = 0. If the number on the other side of the = sign is not 0, then there are far too many choices, for example ( )( ) = 12 means you could have ( ) = 1 & ( ) = 12, or ( ) = 2 & ( ) = 6, or ( ) = ½ & ( ) = 24, etc. Example 3: Solve x 2 + 48 = 14x a) x 2 + 48 = 14x becomes x 2 – 14x + 48 = 0 b) Factoring x 2 – 14x + 48 gives you (x - 6)(x - 8) = 0 c) the Zero -Factor principle set each set of ( ) equal to 0 (x - 6) = 0 or (x - 8) = 0 Solving gives you x = 6 or x = 8 Answer: x = 6, 8 NOTE: When we find the answer for “x”, the answer is also called “the roots”, i.e. “find the roots of this polynomial equation”. X-intercepts So, how is this helpful??? If you tried to graph the equation, 2x 2 + 11x + 15 = y, you would get a parabola, i.e. the U-shaped graph (more on parabolas in Math M-107). When we substitute in “0” for “y”, i.e. finding the roots, you are actually finding where the parabola crosses the x-axis (where the y-coordinate is always 0), i.e. you are finding the x-intercepts! So, for 2x 2 + 11x + 15 = 0, we found that x = -3, -5/2. This mean the parabolas crosses the x-axis (i.e. the x- intercepts) at (-3, 0) and (-5/2, 0).
NOTE: the x-intercepts are also known as the “roots” Extending Example 3: Find the x-intercepts for x 2 + 48 = 14x a) x 2 + 48 = 14x becomes x 2 – 14x + 48 = 0 b) Factoring x 2 – 14x + 48 gives you (x - 6)(x - 8) = 0 c) the Zero -Factor principle set each set of ( ) equal to 0 (x - 6) = 0 or (x - 8) = 0 Solving gives you x = 6 or x = 8 Answer: x-intercepts = (6, 0) & (8, 0) Example 4: Find the x-intercepts for x 2 = 64 d) x 2 = 64 becomes x 2 – 64 = 0 the difference of squares e) Factoring x 2 – 64 gives you (x + 8)(x - 8) = 0 f) the Zero -Factor principle set each set of ( ) equal to 0 (x + 8) = 0 or (x - 8) = 0 Solving gives you x = - 8 or x = 8 Answer: x-intercepts = (-8, 0) & (8, 0) Example 5: 5x 2 – 140 = 15x Set the equation = to 0 5x 2 – 140 = 15x becomes 5x 2 – 15x - 140 = 0 When we factored this polynomial, we got 5(x - 7)(x + 4), so that makes the equation: 5(x - 7)(x + 4) = 0 So, that means either 5 = 0 or (x - 7) = 0 or (x + 4) = 0 So, solve each of them: 5 = 0 or (x - 7) = 0 or (x + 4) = 0 5 cannot = 0, so ignore this part x = 7 or x = -4 so our final answer is: x = 7, -4 So, for 5x 2 – 140 = 15x, we found that x = 7, -4. This mean the parabolas crosses the x- axis at (7, 0) and (-4, 0)
NOTE: When you pull out a GCF, it may or may not be part of your answer. If the GCF has “x” in it, it becomes part of the answer. Example 6: 2x 3 = 18x 2 +72x Set the equation = to 0 2x 3 = 18x 2 +72x becomes 2x 3 - 18x 2 -72x = 0 When we factored this polynomial, we got 2x(x - 12)(x + 3), so that makes the equation: 2x(x - 12)(x + 3) = 0 So, that means either 2x = 0 or (x - 12) = 0 or (x + 3) = 0 So, solve each of them: 2x = 0 or (x - 12) = 0 or (x + 3) = 0 x = 0 or x = 12 or x = -3 so our final answer is: x = 0, 12, -3 So, for 2x 3 = 18x 2 +72x, we found that x = 0, 12, -3. This mean the graph crosses the x- axis at (0, 0), (12, 0) and (-3, 0) This method also works with the Special Cases of the “ac” method Special Case #1: “Perfect Squares” Example 1: Solve x 2 + 12x = - 36 a) x 2 + 12x = - 36 becomes x 2 + 12x + 36 = 0 b) Factoring x 2 + 12x + 36 gives you (x + 6)(x + 6) = 0 c) the Zero -Factor principle set each set of ( ) equal to 0 (x + 6) = 0 or (x + 6) = 0 Solving gives you x = - 6 or x = - 6 Answer: x = - 6 no need to repeat the – 6. This mean the parabolas crosses the x-axis at (-6, 0) only. This happens when only the very bottom point of the parabola touches the x-axis. Example 2: Find the roots for x 2 = 14x - 49
a) x 2 = 14x - 49 becomes x 2 - 14x + 49 = 0 b) Factoring x 2 - 14x + 49 gives you (x - 7)(x - 7) = 0 c) the Zero -Factor principle set each set of ( ) equal to 0 (x - 7) = 0 or (x - 7) = 0 Solving gives you x = 7 or x = 7 Answer: x = 7 no need to repeat the 7 Example 3: x 2 = 81 Set the equation = to 0 x 2 = 81 becomes x 2 - 81 = 0 When we factored this polynomial, we got (x - 9)(x + 9), so that makes the equation: (x - 9)(x + 9) = 0 So, that means either (x - 9) = 0 or (x + 9) = 0 So, solve each of them: (x - 9) = 0 or (x + 9) = 0 x = 9 or x = -9 so our final answer is: x = 9, -9 So, for x 2 = 81, we found that x = 9, -9. This mean the parabolas crosses the x-axis at (9, 0) and (-9, 0) Example 4: Find the roots for x 2 = – 81 x 2 = - 81 becomes x 2 + 81 = 0 x 2 + 81 = 0 does not factor it is prime no roots, i.e. no x-intercepts this graph does not cross the x-axis (it is actually above the x-axis)
Sec. 6.6 Detailed Notes - Week 11 - Linear vs. Quadratic Equations
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