Solving%20for%20x%20by%20Families

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
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 variable

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 -8 -3
  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,
  (x2)
• Isolate the x2 term
• Take the square root of each side
• Remember to include the
• Example 2x2 4 20
• Adding 4 to each side gives 2x2 24
• Dividing each side by 2 gives x2 12
• Take square root of each side gives x 2v3
• Check by substituting 2v3 into the original
  equation

7
Degree ½ (Square Root), no Degree 1 and constants

• If the problem has variables with only a square
  root, (vx)
• Isolate the vx term
• Square each side
• Solve for the variable
• Example 2v(x6) 5 1
• Adding 5 to each side gives 2v(x6) 6
• Dividing each side by 2 gives v(x6) 3
• Square each side gives x6 9
• Subtract 6 from each side gives x 3
• 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 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 4x
• Subtracting 4x gives 3x2 4x 0
• Factoring out the x gives x(3x-4) 0
• 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 6x2 18x 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, constantsand 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 0
• If the leading coefficient (a) equals 1, find the
  factors of c that add to b.

11
Degree 2, Degree 1, constantsand a 1

• Example x2 x 12
• Subtracting 12 from each side gives x2 x 12
  0
• The factors of -12 are (1, -12), (2, -6), (3,
  -4), (4, -3), (6, -2), (12, -1)
• -4 3 -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 constantsand 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 0
• If 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 x
• Factor by grouping.

13
Degree 2, Degree 1 and constantsand a ? 1

• Example 6x2 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
• 6x2 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
• ax2 bx c 0

15
If the quadratic is not factorable

• Example 5x2 4x 4 0
• Using the quadratic equation
• So the roots are imaginary
• x -0.4 0.8i or x -0.4 0.8i