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Grade 9 – Mathematics
Quarter I
SOLVING QUADRATIC EQUATIONS BY
EXTRACTING SQUARE ROOTS
Objectives:
1. familiarize numbers that are
perfect squares; and
2. solve quadratic equations by
extracting square roots.
Quadratic Equations that can be written in the form 𝒙 𝟐
= π’Œ can be solved
by applying the following properties:
1. If π‘˜ > 0, then 𝒙 𝟐 = π’Œ has two real solutions or roots: π‘₯ = Β± π‘˜.
2. If π‘˜ = 0, then 𝒙 𝟐
= π’Œ has two real solutions or roots: π‘₯ = 0
3. If π‘˜ < 0, then 𝒙 𝟐 = π’Œ has no solutions or roots:
Speaking Mathematically
π‘₯ = Β± 36 is read as β€˜π’™ is equal to the positive or negative square root of 36’
π‘₯ βˆ’ 5 2is read as β€œthe square root of the square of the quantity 𝒙 π’Žπ’Šπ’π’–π’” πŸ“".
π‘₯2
= 49
π‘₯2 = Β± 49
π‘₯ = Β± 7
How to extract square roots?
π‘₯2
= 169
π‘₯2 = Β± 169
π‘₯ = Β± 13
How to extract square roots?
8 = 4 βˆ™ 2 32 = 16 βˆ™ 2 54 = 9 βˆ™ 6 75 = 25 βˆ™ 3
12 = 4 βˆ™ 3 40 = 4 βˆ™ 10 56 = 4 βˆ™ 14 76 = 4 βˆ™ 19
18 = 9 βˆ™ 2 44 = 4 βˆ™ 11 54 = 9 βˆ™ 6 80 = 16 βˆ™ 5
20 = 4 βˆ™ 5 45 = 9 βˆ™ 5 60 = 4 βˆ™ 15 90 = 9 βˆ™ 10
24 = 4 βˆ™ 6 48 = 16 βˆ™ 3 63 = 9 βˆ™ 7 96 = 16 βˆ™ 6
27 = 9 βˆ™ 3 50 = 25 βˆ™ 2 68 = 4 βˆ™ 17 98 = 49 βˆ™ 2
28 = 4 βˆ™ 7 52 = 4 βˆ™ 13 72 = 36 βˆ™ 2 99 = 9 βˆ™ 11
8 = 4 βˆ™ 2 54 = 9 βˆ™ 6
12 = 4 βˆ™ 3 56 = 4 βˆ™ 14
18 = 9 βˆ™ 2 54 = 9 βˆ™ 6
20 = 4 βˆ™ 5 60 = 4 βˆ™ 15
24 = 4 βˆ™ 6 63 = 9 βˆ™ 7
27 = 9 βˆ™ 3 68 = 4 βˆ™ 17
28 = 4 βˆ™ 7 72 = 36 βˆ™ 2
32 = 16 βˆ™ 2 75 = 25 βˆ™ 3
40 = 4 βˆ™ 10 76 = 4 βˆ™ 19
44 = 4 βˆ™ 11 80 = 16 βˆ™ 5
45 = 9 βˆ™ 5 90 = 9 βˆ™ 10
48 = 16 βˆ™ 3 96 = 16 βˆ™ 6
50 = 25 βˆ™ 2 98 = 49 βˆ™ 2
52 = 4 βˆ™ 13 99 = 9 βˆ™ 11
π‘₯2
= 75
Get the square root of both sides
π‘₯2 = 75
Factor the perfect squares π‘₯ = Β± 25 βˆ™ 3
Get the square root of the
perfect square. π‘₯ = Β± 25 3
π‘₯ = Β± 5 3
𝟐 𝒙 βˆ’ πŸ“ 𝟐
= πŸ‘πŸ
Divide both sides by 2
𝒙 βˆ’ πŸ“ 𝟐
= πŸπŸ”
Get the square root of both sides 𝒙 βˆ’ πŸ“ 𝟐 = πŸπŸ”
𝒙 βˆ’ πŸ“ = Β± πŸ’
𝒙 βˆ’ πŸ“ = πŸ’
𝒙 = πŸ’ + πŸ“
𝒙 = πŸ—
𝒙 βˆ’ πŸ“ = βˆ’πŸ’
𝒙 = βˆ’πŸ’ + πŸ“
𝒙 = 𝟏
Find the solutions or roots.
𝟐 𝒙 βˆ’ πŸ“ 𝟐
= πŸ‘πŸ
𝟐
πŸ‘ πŸ’π’™ βˆ’ 𝟏 𝟐
βˆ’ 𝟏 = 𝟏𝟏
Divide both sides by 3 πŸ‘ πŸ’π’™ βˆ’ 𝟏 𝟐
= 𝟏𝟐
Get the square root of both sides
πŸ’π’™ βˆ’ 𝟏 𝟐
= πŸ’
πŸ’π’™ βˆ’ 𝟏 = Β± πŸπŸ’π’™ βˆ’ 𝟏 = 𝟐
4𝒙 = πŸ‘
𝒙 =
πŸ‘
πŸ’
πŸ’π’™ βˆ’ 𝟏 = βˆ’πŸ
4𝒙 = βˆ’πŸ
𝒙 =
βˆ’πŸ
πŸ’
Find the solutions or roots.
πŸ’π’™ βˆ’ 𝟏 𝟐 = πŸ’
πŸ‘ πŸ’π’™ βˆ’ 𝟏 𝟐
= 𝟏𝟐
πŸ‘
πŸπ’™ βˆ’ πŸ‘ 𝟐
= πŸπŸ–
Get the square root of both sides
2π‘₯ βˆ’ 3 2 = 18
2π‘₯ βˆ’ 3 2 = Β± 9 βˆ™ 2
2π‘₯ βˆ’ 3 = Β± 9 2
2π‘₯ βˆ’ 3 = Β±3 2
2π‘₯ βˆ’ 3 = Β±3 2
πŸπ’™ βˆ’ πŸ‘ = πŸ‘ 𝟐
πŸπ’™ = πŸ‘ + πŸ‘ 𝟐
𝒙 =
πŸ‘ + πŸ‘ 𝟐
𝟐
πŸπ’™ βˆ’ πŸ‘ = βˆ’πŸ‘ 𝟐
πŸπ’™ = πŸ‘ βˆ’ πŸ‘ 𝟐
𝒙 =
πŸ‘ βˆ’ πŸ‘ 𝟐
𝟐
𝟐 πŸ“π’™ + 𝟐 𝟐
= πŸ”πŸ’
Get the square root of both sides
5π‘₯ + 2 2 = 32
5π‘₯ + 2 2 = Β± 16 βˆ™ 2
5π‘₯ + 2 = Β± 16 2
5π‘₯ + 2 = Β±4 2
5π‘₯ + 2 = Β± 4 2
πŸ“π’™ + 𝟐 = πŸ’ 𝟐
πŸ“π’™ = βˆ’πŸ + πŸ’ 𝟐
𝒙 =
βˆ’πŸ + πŸ’ 𝟐
πŸ“
Divide both sides by 2
5π‘₯ + 2 2
= 32
πŸ“π’™ + 𝟐 = βˆ’ πŸ’ 𝟐
πŸ“π’™ = βˆ’πŸ βˆ’ πŸ’ 𝟐
𝒙 =
βˆ’πŸ βˆ’ πŸ’ 𝟐
πŸ“
𝟐 πŸ“π’™ + 𝟐 𝟐
= πŸ”πŸ’
𝟐

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Solving Quadratic Equations by Extracting Square Roots

  • 1. Grade 9 – Mathematics Quarter I SOLVING QUADRATIC EQUATIONS BY EXTRACTING SQUARE ROOTS
  • 2. Objectives: 1. familiarize numbers that are perfect squares; and 2. solve quadratic equations by extracting square roots.
  • 3.
  • 4. Quadratic Equations that can be written in the form 𝒙 𝟐 = π’Œ can be solved by applying the following properties: 1. If π‘˜ > 0, then 𝒙 𝟐 = π’Œ has two real solutions or roots: π‘₯ = Β± π‘˜. 2. If π‘˜ = 0, then 𝒙 𝟐 = π’Œ has two real solutions or roots: π‘₯ = 0 3. If π‘˜ < 0, then 𝒙 𝟐 = π’Œ has no solutions or roots: Speaking Mathematically π‘₯ = Β± 36 is read as β€˜π’™ is equal to the positive or negative square root of 36’ π‘₯ βˆ’ 5 2is read as β€œthe square root of the square of the quantity 𝒙 π’Žπ’Šπ’π’–π’” πŸ“".
  • 5. π‘₯2 = 49 π‘₯2 = Β± 49 π‘₯ = Β± 7 How to extract square roots?
  • 6. π‘₯2 = 169 π‘₯2 = Β± 169 π‘₯ = Β± 13 How to extract square roots?
  • 7. 8 = 4 βˆ™ 2 32 = 16 βˆ™ 2 54 = 9 βˆ™ 6 75 = 25 βˆ™ 3 12 = 4 βˆ™ 3 40 = 4 βˆ™ 10 56 = 4 βˆ™ 14 76 = 4 βˆ™ 19 18 = 9 βˆ™ 2 44 = 4 βˆ™ 11 54 = 9 βˆ™ 6 80 = 16 βˆ™ 5 20 = 4 βˆ™ 5 45 = 9 βˆ™ 5 60 = 4 βˆ™ 15 90 = 9 βˆ™ 10 24 = 4 βˆ™ 6 48 = 16 βˆ™ 3 63 = 9 βˆ™ 7 96 = 16 βˆ™ 6 27 = 9 βˆ™ 3 50 = 25 βˆ™ 2 68 = 4 βˆ™ 17 98 = 49 βˆ™ 2 28 = 4 βˆ™ 7 52 = 4 βˆ™ 13 72 = 36 βˆ™ 2 99 = 9 βˆ™ 11
  • 8. 8 = 4 βˆ™ 2 54 = 9 βˆ™ 6 12 = 4 βˆ™ 3 56 = 4 βˆ™ 14 18 = 9 βˆ™ 2 54 = 9 βˆ™ 6 20 = 4 βˆ™ 5 60 = 4 βˆ™ 15 24 = 4 βˆ™ 6 63 = 9 βˆ™ 7 27 = 9 βˆ™ 3 68 = 4 βˆ™ 17 28 = 4 βˆ™ 7 72 = 36 βˆ™ 2 32 = 16 βˆ™ 2 75 = 25 βˆ™ 3 40 = 4 βˆ™ 10 76 = 4 βˆ™ 19 44 = 4 βˆ™ 11 80 = 16 βˆ™ 5 45 = 9 βˆ™ 5 90 = 9 βˆ™ 10 48 = 16 βˆ™ 3 96 = 16 βˆ™ 6 50 = 25 βˆ™ 2 98 = 49 βˆ™ 2 52 = 4 βˆ™ 13 99 = 9 βˆ™ 11 π‘₯2 = 75 Get the square root of both sides π‘₯2 = 75 Factor the perfect squares π‘₯ = Β± 25 βˆ™ 3 Get the square root of the perfect square. π‘₯ = Β± 25 3 π‘₯ = Β± 5 3
  • 9. 𝟐 𝒙 βˆ’ πŸ“ 𝟐 = πŸ‘πŸ Divide both sides by 2 𝒙 βˆ’ πŸ“ 𝟐 = πŸπŸ” Get the square root of both sides 𝒙 βˆ’ πŸ“ 𝟐 = πŸπŸ” 𝒙 βˆ’ πŸ“ = Β± πŸ’ 𝒙 βˆ’ πŸ“ = πŸ’ 𝒙 = πŸ’ + πŸ“ 𝒙 = πŸ— 𝒙 βˆ’ πŸ“ = βˆ’πŸ’ 𝒙 = βˆ’πŸ’ + πŸ“ 𝒙 = 𝟏 Find the solutions or roots. 𝟐 𝒙 βˆ’ πŸ“ 𝟐 = πŸ‘πŸ 𝟐
  • 10. πŸ‘ πŸ’π’™ βˆ’ 𝟏 𝟐 βˆ’ 𝟏 = 𝟏𝟏 Divide both sides by 3 πŸ‘ πŸ’π’™ βˆ’ 𝟏 𝟐 = 𝟏𝟐 Get the square root of both sides πŸ’π’™ βˆ’ 𝟏 𝟐 = πŸ’ πŸ’π’™ βˆ’ 𝟏 = Β± πŸπŸ’π’™ βˆ’ 𝟏 = 𝟐 4𝒙 = πŸ‘ 𝒙 = πŸ‘ πŸ’ πŸ’π’™ βˆ’ 𝟏 = βˆ’πŸ 4𝒙 = βˆ’πŸ 𝒙 = βˆ’πŸ πŸ’ Find the solutions or roots. πŸ’π’™ βˆ’ 𝟏 𝟐 = πŸ’ πŸ‘ πŸ’π’™ βˆ’ 𝟏 𝟐 = 𝟏𝟐 πŸ‘
  • 11. πŸπ’™ βˆ’ πŸ‘ 𝟐 = πŸπŸ– Get the square root of both sides 2π‘₯ βˆ’ 3 2 = 18 2π‘₯ βˆ’ 3 2 = Β± 9 βˆ™ 2 2π‘₯ βˆ’ 3 = Β± 9 2 2π‘₯ βˆ’ 3 = Β±3 2 2π‘₯ βˆ’ 3 = Β±3 2 πŸπ’™ βˆ’ πŸ‘ = πŸ‘ 𝟐 πŸπ’™ = πŸ‘ + πŸ‘ 𝟐 𝒙 = πŸ‘ + πŸ‘ 𝟐 𝟐 πŸπ’™ βˆ’ πŸ‘ = βˆ’πŸ‘ 𝟐 πŸπ’™ = πŸ‘ βˆ’ πŸ‘ 𝟐 𝒙 = πŸ‘ βˆ’ πŸ‘ 𝟐 𝟐
  • 12. 𝟐 πŸ“π’™ + 𝟐 𝟐 = πŸ”πŸ’ Get the square root of both sides 5π‘₯ + 2 2 = 32 5π‘₯ + 2 2 = Β± 16 βˆ™ 2 5π‘₯ + 2 = Β± 16 2 5π‘₯ + 2 = Β±4 2 5π‘₯ + 2 = Β± 4 2 πŸ“π’™ + 𝟐 = πŸ’ 𝟐 πŸ“π’™ = βˆ’πŸ + πŸ’ 𝟐 𝒙 = βˆ’πŸ + πŸ’ 𝟐 πŸ“ Divide both sides by 2 5π‘₯ + 2 2 = 32 πŸ“π’™ + 𝟐 = βˆ’ πŸ’ 𝟐 πŸ“π’™ = βˆ’πŸ βˆ’ πŸ’ 𝟐 𝒙 = βˆ’πŸ βˆ’ πŸ’ 𝟐 πŸ“ 𝟐 πŸ“π’™ + 𝟐 𝟐 = πŸ”πŸ’ 𝟐