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Identifying “families of problems” is a great way of approaching problems. Problems within the same family are solved with the same process. This PowerPoint.

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Presentation on theme: "Identifying “families of problems” is a great way of approaching problems. Problems within the same family are solved with the same process. This PowerPoint."— Presentation transcript:

1 Identifying “families of problems” is a great way of approaching problems. Problems within the same family are solved with the same process. This PowerPoint identifies families of problems in solving for a variable. We will be using the variable “x”, but the variable can be any letter.

2 PEMDAS Students learn the acronym PEMDAS to help them remember the order of operations in solving mathematical problems. The letters stand for: P = Parenthesis E = Exponents M = Multiplication D = Division A = Addition S = Subtraction

3 PEMDAS Students use the order of operation from left to right to simplify expressions, i.e., simplifying each side of the equation prior to working across the equality sign. Students use the order from right to left to solve for the variable. Solving for the variable can also be thought of as “peeling the onion” where the student works from the outside to unwrap the problem and solve for the variable.

4 Simplifying an expression PEMDAS Solving the equation for the varable

5 Degree 1 and constant If the degree of the x terms are only single order, simply solve for x using PEMDAS Simplify each side by using PEMDAS from left to right Then solve for x by using PEMDAS from right to left Example: 2x + 2(x -4) – 3 = 11 Simplifying by distributing gives 2x + 2x = 11 Simplifying by gathering like terms: 4x – 11 = 11 Solving by adding 11 to each side gives 4x = 22 Dividing each side by 4 gives x = 5.5 Check by substituting x = 5.5 into the original problem

6 Degree 2, no Degree 1 and constants If the problem has variables with only degree 2, (x 2 ): 1. Isolate the x 2 term 2. Take the square root of each side 3. Remember to include the ± 4. Example: 2x 2 – 4 = Adding 4 to each side gives: 2x 2 = Dividing each side by 2 gives: x 2 = Take square root of each side gives: x = ±2√3 8. Check by substituting ±2√3 into the original equation

7 Degree ½ (Square Root), no Degree 1 and constants If the problem has variables with only a square root, (√x): 1. Isolate the √x term 2. Square each side 3. Solve for the variable 4. Example: 2√(x+6) – 5 = 1 5. Adding 5 to each side gives: 2√(x+6) = 6 6. Dividing each side by 2 gives: √(x+6) = 3 7. Square each side gives: x+6 = 9 8. Subtract 6 from each side gives: x = 3 9. Check by substituting 3 into the original equation

8 Degree 2, Degree 1 and no constants If the problem has variables with degree 1 and 2, (x and x 2 ), but no constants: 1. Move everything to one side of the equal sign. 2. Factor out a common x. 3. Use the zero product rule to determine the solutions. 4. Example: 3x 2 = 4x 5. Subtracting 4x gives: 3x 2 – 4x = 0 6. Factoring out the x gives: x(3x-4) = 0 7. Either x = 0 or 3x-4 = 0, so x = 0 or x = 4/3

9 Four terms If the equation has four terms, try factoring by grouping and using the zero product rule: Example: 6x x – 8x – 24 = 0 Find the common factor of the first two and last two terms: 6x(x + 3) – 8(x + 3) = 0 Take out the common factor of (x + 3): (x + 3)(6x – 8) = 0 Use the zero product rule: x + 3 = 0 or 6x – 8 = 0 x = -3 or x = 4/3

10 Degree 2, Degree 1, constants and “a” = 1 If the problem has variables with degree 1 and 2, (x and x 2 ) as well as constants: 1. Move everything to one side of the equal sign, making everything equal zero, leaving the equation in standard form of ax 2 + bx + c = 0 2. If the leading coefficient (a) equals 1, find the factors of c that add to b.

11 Degree 2, Degree 1, constants and “a” = 1 Example: x 2 – x = 12 Subtracting 12 from each side gives: x 2 – x – 12 = 0 The factors of -12 are: (1, -12), (2, -6), (3, -4), (4, -3), (6, - 2), (12, -1) = -1, so the factors are: (x – 4) * (x + 3) = 0 Using the zero product rule gives: x– 4 = 0 or x + 3 = 0, so x = 4 or x = -3 Check by substituting 4 and -3 into the original equation

12 Degree 2, Degree 1 and constants and “a” ≠ 1 If the problem has variables with degree 1 and 2, (x and x 2 ): 1. Move everything to one side of the equal sign, making everything equal zero, leaving the equation in standard form of ax 2 + bx + c = 0 2. If the leading coefficient (a) does not equal 1, find the factors of a * c that add to b. 3. Replace the “bx” term with the two factors multiplied by x 4. Factor by grouping.

13 Degree 2, Degree 1 and constants and “a” ≠ 1 Example: 6x 2 – 7x – 5 = 0 a * c = 6 * (-5) = -30, the factors of -30 are: (1, -30), (2, -15), (3, -10), (5, -6), (6, -5), (10, -3), (15, -2) and (30, -1) 3 – 10 = -7, so substitute 3x – 10x for -7x, giving: 6x 2 + 3x – 10x – 5 = 0 Factor by grouping: 3x(2x + 1) – 5(2x + 1) = 0 (2x + 1) * (3x – 5) = 0 Using the zero product rule gives 2x + 1 = 0 or 3x – 5 = 0, so x = -1/5 or x = 5/3 Check by substituting -1/5 and 5/3 into the original equation

14 If the quadratic is not factorable The above techniques only work if the roots (zeros, solutions, x-intercepts) are real, and rational. They do not work for imaginary roots, and may not work for radical roots. The quadratic formula can be used to find the factors of any quadratic equation. The roots can be found by: Where the a, b, and c are from the standard form ax 2 + bx + c = 0

15 If the quadratic is not factorable Example: 5x 2 + 4x +4 = 0 Using the quadratic equation: So the roots are imaginary: x = i or x = -0.4 – 0.8i


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