5/3/17 Week 5 Wednesday CHALLENGE daily sheet:
Solving quadratic equations by completing the square
Daily aims:
 1. I can form perfect square trinomials.
 2. I can solve quadratic equations by completing the square when a = 1 and b is even.
Before lesson
1) What’s the standard form of a quadratic equation?
During lesson
2a) (x + 4)2 = (x + 4)(x + 4) =
2b) (x – 3)2 =
Find the value you need to add in order to form a perfect
square trinomial in each case. (Divide b—the
coefficient of x— by 2 and square the result.)
3a) x2 + 8x + _____
3b) x2 – 6x + _____
D. Stark 3/25/2017
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Solve the quadratic equations by completing the square.
1. Put the constant on the right [to get it out of the way.]
2. Add to both sides what’s needed to form a
perfect square trinomial.
3. Factor the perfect square trinomial.
4. Use the method of finding roots to finish.
4a) x2 + 8x – 9 = 0
4b) x2 – 6x – 27 = 0
D. Stark 3/25/2017
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5/3/17 Week 5 Wednesday CHALLENGE daily sheet: Solving quadratic equations by completing the square
KEY
Before lesson
1) What’s the standard form of a quadratic equation?
During lesson
ax2 ± bx ± c = 0
This is what I’m talking about when I speak of a = 1
or a > 1 or b is even/odd—the coefficients of the x2
and x terms.
2a) (x + 4)2 = (x + 4)(x + 4)
= x2 + 4x + 4x + 16
= x2 + 8x + 16
2b) (x – 3)2 =
= (x – 3)(x – 3)
= x2 – 3x – 3x + 9
= x2 – 6x + 9
The results here are perfect square trinomials.
They are trinomials that can be factored into the
square of a simple sum or difference.
Note that if you divide the coefficient of x by 2 and
then square the result, you’ll get the constant term.
D. Stark 3/25/2017
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Find the value you need to add in order to form a perfect square
trinomial in each case. (Divide b—the coefficient of x— by 2 and
square the result.)
3a) x2 + 8x + 16
Divide 8 by 2 and square the result.
3b) x2 – 6x + 9
Divide –6 by 2 and square the result.
Solve the quadratic equations by completing the square.
1. Put the constant on the right [to get it out of the way.]
2. Add to both sides what’s needed to form a perfect square
trinomial.
3. Factor the perfect square trinomial.
4. Use the method of finding roots to finish.
4a) x2 + 8x – 9 = 0
x2 + 8x
=9
4b) x2 – 6x – 27 = 0
x2 – 6x
= 27
x2 + 8x + 16 = 9 + 16
(x + 4)2 = 9 + 16
(x + 4)2 = 25
x2 – 6x + 9 = 27 + 9
(x – 3)2 = 27 + 9
(x – 3)2 = 36
√(𝒙 + 𝟒)𝟐 = ±√𝟐𝟓
√(𝒙 – 𝟑)𝟐 = ±√𝟑𝟔
x+4= ±5
x + 4 = 5 or x + 4 = –5
x = 1 or x = –9
solution set: {–9, 1}
x–3= ±6
x – 3 = 6 or x – 3 = –6
x = 9 or x = –3
solution set: {–3, 9}
D. Stark 3/25/2017
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