Quadratics Day 3

(1)

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The Quadratic Formula and the

(2)

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The Quadratic Formula and the

Discriminant

Solve

quadratic equations by

Factoring and the Quadratic Formula.

(3)

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The Quadratic Formula and the

Discriminant

Teacher Example: Solving Quadratic Equations by Factoring

Solve the quadratic equation by factoring. Check your answer.

x2 + 4x = 21

x2 + 4x = 21 –21 –21

x2 + 4x – 21 = 0 (x + 7)(x –3) = 0

x + 7 = 0 or x – 3 = 0

x = –7 or x = 3

The solutions are –7 and 3.

The equation must be written in standard form. So subtract 21 from both sides.

Factor the trinomial.

(4)

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The Quadratic Formula and the

Discriminant

• We can check these solutions by plugging

them in and when solved, the answer

(5)

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The Quadratic Formula and the

Discriminant

Student Example 1

x2 + 4x = 5

x2 + 4x = 5 –5 –5

x2 + 4x – 5 = 0

Write the equation in standard form. Add – 5 to both sides.

Use the Zero Product Property.

Solve each equation.

(x – 1)(x + 5) = 0

x – 1 = 0 or x + 5 = 0

x = 1 or x = –5

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The Quadratic Formula and the

Discriminant

Student Example 2

x2 – 12x + 36 = 0 (x – 6)(x – 6) = 0

x – 6 = 0 or x – 6 = 0

x = 6 or x = 6

Both factors result in the same solution, so there is one solution, 6.

Use the Zero Product Property.

(7)

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The Quadratic Formula and the

Discriminant

Teacher Example : Solving Quadratic Equations by Factoring

–2x2 = 20x + 50

The equation must be written in standard form. So add 2x2 to

both sides.

Factor out the GCF 2.

+2x2 +2x2

0 = 2x2 + 20x + 50

–2x2 = 20x + 50

2x2 + 20x + 50 = 0 2(x2 + 10x + 25) = 0

2(x + 5)(x + 5) = 0 2 ≠ 0 or x + 5 = 0

x = –5

(8)

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The Quadratic Formula and the

Discriminant

Teacher Example Continued

–2x2 = 20x + 50

Check

–2x2 = 20x + 50

–2(–5)2 20(–5) + 50

–50 –100 + 50 –50 –50

_

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The Quadratic Formula and the

Discriminant

Student Example 3

30x = –9x2 – 25

–9x2 – 30x – 25 = 0

–1(3x + 5)(3x + 5) = 0

–1(9x2 + 30x + 25) = 0

–1 ≠ 0 or 3x + 5 = 0

Write the equation in standard form.

Use the Zero Product Property. – 1 cannot equal 0.

(10)

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The Quadratic Formula and the

(11)

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The Quadratic Formula and the

Discriminant

Teacher Example: Using the Quadratic Formula

Solve using the Quadratic Formula. 6x2 + 5x – 4 = 0

6x2 + 5x + (–4) = 0

Identify a, b, and c.

Use the Quadratic Formula.

Simplify.

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The Quadratic Formula and the

Discriminant

Solve using the Quadratic Formula. 6x2 + 5x – 4 = 0

Simplify.

Write as two equations.

(13)

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The Quadratic Formula and the

Discriminant

Teacher Example 2: Using the Quadratic Formula

Solve using the Quadratic Formula.

x2 = x + 20

1x2 + (–1x) + (–20) = 0 Write in standard form. Identify

a, b, and c.

Use the quadratic formula.

Simplify.

(14)

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The Quadratic Formula and the

Discriminant

Teacher Example 2 Continued

x = 5 or x = –4

Simplify.

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The Quadratic Formula and the

Discriminant

Student Example 1

–3x2 + 5x + 2 = 0

Identify a, b, and c.

Substitute –3 for a, 5 for b, and 2 for c.

Simplify

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The Quadratic Formula and the

Discriminant

Student Example 1 Continued

Simplify.

(17)

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The Quadratic Formula and the

Discriminant

Student Example 2

Solve using the Quadratic Formula. 2 – 5x2 = –9x

Write in standard form. Identify a, b, and c.

(–5)x2 + 9x + (2) = 0

Substitute –5 for a, 9 for b, and 2 for c.

(18)

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The Quadratic Formula and the

Discriminant

Student Example 2 Continued

Simplify.

2 – 5x2 = –9x

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The Quadratic Formula and the

Discriminant

Many quadratic equations can be solved by

graphing, factoring, taking the square root, or

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The Quadratic Formula and the

Discriminant

Teacher Example: Application

The height in feet of a diver above the water can be modeled by h(t) = –16t2 + 8t + 8, where t is

time in seconds after the diver jumps off a

platform. Find the time it takes for the diver to reach the water.

h = –16t2 + 8t + 8

0 = –16t2 + 8t + 8

0 = –8(2t2 – t – 1)

0 = –8(2t + 1)(t – 1)

The diver reaches the water when h = 0.

(21)

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The Quadratic Formula and the

Discriminant

–8 ≠ 0, 2t + 1 = 0 or t – 1= 0 Use the Zero Product _Property.

2t = –1 or t = 1 _{Solve each equation.}

It takes the diver 1 second to reach the water.

Check 0 = –16t2 + 8t + 8

Substitute 1 into the original equation.

0 –16(1)2 + 8(1) + 8

0 –16 + 8 + 8 0 0

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Since time cannot be

negative, does not make sense in this

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The Quadratic Formula and the

Discriminant

Student Example

What if…? The equation for the height above the water for another diver can be modeled by h = – 16t2 + 8t + 24. Find the time it takes this diver to

reach the water.

h = –16t2 + 8t + 24

0 = –16t2 + 8t + 24

0 = –8(2t2 – t – 3)

The diver reaches the water when h = 0.

(23)

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The Quadratic Formula and the

Discriminant

–8 ≠ 0, 2t – 3 = 0 or t + 1= 0 Use the Zero Product

Property.

2t = 3 or t = –1 _{Solve each equation.}

Since time cannot be negative, –1 does not make sense in this situation.

It takes the diver 1.5 seconds to reach the water.

Check 0 = –16t2 + 8t + 24

Substitute 1 into the original equation.

0 –16(1.5)2 + 8(1.5) + 24

0 –36 + 12 + 24 0 0

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Student Example Continued

t = 1.5

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The Quadratic Formula and the

Discriminant

Exit Ticket Part 2

Solve each quadratic equation by factoring. Check your answer.

1. x2 + 16x + 48 = 0

–4, –12

2. –4x2 = 16x + 16 –2

3. The height of a rocket launched upward from a 160 foot cliff