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Section 9.4 Day 1 Solving Quadratic Equations by Completing the Square

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Presentation on theme: "Section 9.4 Day 1 Solving Quadratic Equations by Completing the Square"β€” Presentation transcript:

1 Section 9.4 Day 1 Solving Quadratic Equations by Completing the Square
Algebra 1

2 Complete the square to write a perfect square trinomial
Convert standard form to vertex form by completing the square Solve a quadratic equation by using the square root property Solve a quadratic equation by completing the square Learning Targets

3 Completing the Square Procedure
1. Determine π‘Ž & 𝑏 2. Find 𝐢= 𝑏 2π‘Ž 2 3. Write 𝑏 2π‘Ž 2 on both sides of the equation 4. Rewrite into π‘₯+ 𝑏 2π‘Ž 2 Completing the Square Procedure

4 Completing the Square – Example 1
Find the value of 𝑐 that makes π‘₯ 2 +4π‘₯+𝑐 a perfect square trinomial 1. π‘Ž=1, 𝑏=4, 𝑐=? 2. 𝑏 2 = =2 and 𝑏 = 2 2 =4 3. Rewrite: π‘₯+2 2 = π‘₯ 2 +4π‘₯+4 4. 𝑐=4 Completing the Square – Example 1

5 Completing the Square – Example 2
Find the value of 𝑐 that makes π‘Ÿ 2 βˆ’8π‘Ÿ+𝑐 a perfect square trinomial 1. 𝑏 2 = βˆ’ 8 2 =βˆ’4 and 𝑏 =(βˆ’ 4) 2 =16 2. 𝑐=16 3. π‘Ÿ 2 βˆ’8π‘Ÿ+16= π‘Ÿβˆ’4 2 Completing the Square – Example 2

6 Standard to Vertex Form – Example 1
Convert 𝑦= π‘₯ 2 βˆ’6π‘₯+12 into vertex form 1. Complete the square: 𝑏 = βˆ’3 2 =9 2. Add the number to both sides 𝑦+9= π‘₯ 2 βˆ’6π‘₯+9+12 3. Group: 𝑦+9= π‘₯βˆ’ 4. Simplify: 𝑦= π‘₯βˆ’ Standard to Vertex Form – Example 1

7 Standard to Vertex Form – Example 2
Convert 𝑦= π‘₯ 2 βˆ’12π‘₯+3 into vertex form 1. Complete the square: 𝑏 = βˆ’6 2 =36 2. Add the number to both sides 𝑦+36= π‘₯ 2 βˆ’12π‘₯+36+3 3. Group together: 𝑦+36= π‘₯βˆ’ 4. Simplify: 𝑦= π‘₯βˆ’6 2 βˆ’33 Standard to Vertex Form – Example 2

8 Solving using the Square Root Property – Example 1
Solve π‘₯ 2 =16 1. Take the square root of each side 2. π‘₯=Β± 16 3. π‘₯=Β±4 Key Note: The symbol does not represent taking the square root. It represents the positive square root. Solving using the Square Root Property – Example 1

9 Solving using the Square Root Property – Example 2
Solve π‘₯βˆ’6 2 =81 1. Take the square root of both sides π‘₯βˆ’6=Β± 81 2. Simplify and solve: π‘₯βˆ’6=Β±9 π‘₯βˆ’6= and π‘₯βˆ’6=βˆ’9 π‘₯=15 and π‘₯=βˆ’3 Solving using the Square Root Property – Example 2

10 Solving using the Square Root Property – Example 3
Solve π‘₯+3 2 =25 π‘₯+3=Β± 25 π‘₯+3=Β±5 π‘₯+3= and π‘₯+3=βˆ’5 π‘₯=2 and π‘₯=βˆ’8 Solving using the Square Root Property – Example 3

11 Solving by Completing the Square – Example 1
Solve π‘₯ 2 βˆ’12π‘₯+3=16 by completing the square. 1. Complete the square: 𝑏 = βˆ’6 2 =36 2. Add to both sides π‘₯ 2 βˆ’12π‘₯+36+3=16+36 3. Group: π‘₯βˆ’ =52 4. Solve: π‘₯βˆ’6 2 =49 5. π‘₯βˆ’6=Β±7 6. π‘₯=13 and π‘₯=βˆ’1 Solving by Completing the Square – Example 1

12 Solving by Completing the Square – Example 2
Solve π‘₯ 2 +6π‘₯+5=12 by completing the square 1. Complete the square: 𝑏 = =9 2. Add to both sides π‘₯ 2 +6π‘₯+9+5=12+9 π‘₯ =21 3. Solve: π‘₯+3 2 =16 π‘₯+3=Β±4 π‘₯=1 and π‘₯=βˆ’7 Solving by Completing the Square – Example 2


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