Presentation on theme: "Solving for x by Families"— Presentation transcript:
1 Solving for x by Families 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 PEMDASStudents learn the acronym PEMDAS to help them remember the order of operations in solving mathematical problems. The letters stand for:P = ParenthesisE = ExponentsM = MultiplicationD = DivisionA = AdditionS = Subtraction
3 PEMDASStudents 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 PEMDASSolving the equation for the varable
5 Degree 1 and constantIf the degree of the x terms are only single order, simply solve for x using PEMDASSimplify each side by using PEMDAS from left to rightThen solve for x by using PEMDAS from right to leftExample: 2x + 2(x -4) – 3 = 11Simplifying by distributing gives 2x + 2x = 11Simplifying by gathering like terms: 4x – 11 = 11Solving by adding 11 to each side gives 4x = 22Dividing each side by 4 gives x = 5.5Check 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, (x2):Isolate the x2 termTake the square root of each sideRemember to include the ±Example: 2x2 – 4 = 20Adding 4 to each side gives: 2x2 = 24Dividing each side by 2 gives: x2 = 12Take square root of each side gives: x = ±2√3Check 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):Isolate the √x termSquare each sideSolve for the variableExample: 2√(x+6) – 5 = 1Adding 5 to each side gives: 2√(x+6) = 6Dividing each side by 2 gives: √(x+6) = 3Square each side gives: x+6 = 9Subtract 6 from each side gives: x = 3Check 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 x2), but no constants:Move everything to one side of the equal sign.Factor out a common x.Use the zero product rule to determine the solutions.Example: 3x2 = 4xSubtracting 4x gives: 3x2 – 4x = 0Factoring out the x gives: x(3x-4) = 0Either x = 0 or 3x-4 = 0, so x = 0 or x = 4/3
9 Four termsIf the equation has four terms, try factoring by grouping and using the zero product rule:Example: 6x2 + 18x – 8x – 24 = 0Find the common factor of the first two and last two terms: 6x(x + 3) – 8(x + 3) = 0Take out the common factor of (x + 3):(x + 3)(6x – 8) = 0Use the zero product rule:x + 3 = 0 or 6x – 8 = 0x = -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 x2) as well as constants:Move everything to one side of the equal sign, making everything equal zero, leaving the equation in standard form of ax2 + bx + c = 0If 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: x2 – x = 12Subtracting 12 from each side gives: x2 – x – 12 = 0The 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) = 0Using the zero product rule gives:x– 4 = 0 or x + 3 = 0,so x = 4 or x = -3Check 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 x2):Move everything to one side of the equal sign, making everything equal zero, leaving the equation in standard form of ax2 + bx + c = 0If the leading coefficient (a) does not equal 1, find the factors of a * c that add to b.Replace the “bx” term with the two factors multiplied by xFactor by grouping.
13 Degree 2, Degree 1 and constants and “a” ≠ 1 Example: 6x2 – 7x – 5 = 0a * 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:6x2 + 3x – 10x – 5 = 0Factor by grouping:3x(2x + 1) – 5(2x + 1) = 0(2x + 1) * (3x – 5) = 0Using the zero product rule gives2x + 1 = 0 or 3x – 5 = 0, sox = -1/5 or x = 5/3Check 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 formax2 + bx + c = 0
15 If the quadratic is not factorable Example: 5x2 + 4x +4 = 0Using the quadratic equation:So the roots are imaginary:x = i or x = -0.4 – 0.8i