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Review: Find the LCD
1.
𝟏
πŸ‘
+
𝟏
πŸ”
LCD:
2.
πŸ’
𝒙
βˆ’
πŸπ’™βˆ’πŸ
πŸ“
LCD:
3.
πŸπ’™
πŸ‘
+
𝒙+𝟏
𝒙
LCD:
4.
𝟏
𝒙
+
πŸπ’™
πŸ“
LCD:
Transforming
rational algebraic
equation to
quadratic equation
Quadratic Equations
Quadratic Equations are mathematical sentences of
degree 2 that can be written in the form π’‚π’™πŸ
+𝒃𝒙 +
𝒄 = 𝟎 where a, b, and c are real numbers and π‘Ž β‰ 
0
Sometimes quadratic equation are in disguise, to
get the roots of the given rational algebraic equation
transform it to the standard form of ax2+bx+c=0
Example:
Cross multiply
Simplify
Transpose the terms
Arrange the terms to standard form
Multiply the whole equation by -1 to
make the equation positive
3
π‘₯
=
π‘₯+4
7
3 7 = π‘₯ π‘₯ + 4
21 = π‘₯2
+ 4π‘₯
21βˆ’π‘₯2
= 4π‘₯
21 βˆ’π‘₯2
βˆ’4π‘₯ = 0
βˆ’π‘₯2
βˆ’ 4π‘₯ + 21 = 0
(βˆ’π‘₯2
βˆ’4π‘₯ + 21 = 0) (βˆ’1)
π‘₯2
+ 4π‘₯ βˆ’ 21 = 0
Example:
2
𝑑
βˆ’
4𝑑
2
= 7
4 βˆ’ 4𝑑2
2𝑑
= 7
4 βˆ’ 4𝑑2
2𝑑
=
7
1
4 βˆ’ 4𝑑2
1 = 2𝑑(7)
4 βˆ’ 4𝑑2
= 14𝑑
βˆ’4𝑑2
βˆ’ 14𝑑 + 4 = 0
(βˆ’4𝑑2
βˆ’14𝑑 + 4 = 0) (βˆ’1)
4𝑑2
+ 14π‘₯ βˆ’ 4 = 0
Get the LCD of the fractions then subtract
Transpose the terms
Use 1 as the denominator of 7 and then
cross multiply
Multiply the whole equation by -1 to
make the equation positive
Transpose and then arrange to
standard form
S.W. Think-Pair-Share
1.
5
π‘₯
=
π‘₯+3
7
2.
4
𝑏
+
2𝑏
5
= 6

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Grade 9_Week 6_Day 1.pptx

  • 1.
  • 2. Review: Find the LCD 1. 𝟏 πŸ‘ + 𝟏 πŸ” LCD: 2. πŸ’ 𝒙 βˆ’ πŸπ’™βˆ’πŸ πŸ“ LCD: 3. πŸπ’™ πŸ‘ + 𝒙+𝟏 𝒙 LCD: 4. 𝟏 𝒙 + πŸπ’™ πŸ“ LCD:
  • 4. Quadratic Equations Quadratic Equations are mathematical sentences of degree 2 that can be written in the form π’‚π’™πŸ +𝒃𝒙 + 𝒄 = 𝟎 where a, b, and c are real numbers and π‘Ž β‰  0 Sometimes quadratic equation are in disguise, to get the roots of the given rational algebraic equation transform it to the standard form of ax2+bx+c=0
  • 5. Example: Cross multiply Simplify Transpose the terms Arrange the terms to standard form Multiply the whole equation by -1 to make the equation positive 3 π‘₯ = π‘₯+4 7 3 7 = π‘₯ π‘₯ + 4 21 = π‘₯2 + 4π‘₯ 21βˆ’π‘₯2 = 4π‘₯ 21 βˆ’π‘₯2 βˆ’4π‘₯ = 0 βˆ’π‘₯2 βˆ’ 4π‘₯ + 21 = 0 (βˆ’π‘₯2 βˆ’4π‘₯ + 21 = 0) (βˆ’1) π‘₯2 + 4π‘₯ βˆ’ 21 = 0
  • 6. Example: 2 𝑑 βˆ’ 4𝑑 2 = 7 4 βˆ’ 4𝑑2 2𝑑 = 7 4 βˆ’ 4𝑑2 2𝑑 = 7 1 4 βˆ’ 4𝑑2 1 = 2𝑑(7) 4 βˆ’ 4𝑑2 = 14𝑑 βˆ’4𝑑2 βˆ’ 14𝑑 + 4 = 0 (βˆ’4𝑑2 βˆ’14𝑑 + 4 = 0) (βˆ’1) 4𝑑2 + 14π‘₯ βˆ’ 4 = 0 Get the LCD of the fractions then subtract Transpose the terms Use 1 as the denominator of 7 and then cross multiply Multiply the whole equation by -1 to make the equation positive Transpose and then arrange to standard form