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16.2 Finding Solutions to Quadratics

16.2 Finding Solutions to Quadratics

Assessment

Presentation

•

Mathematics

•

9th Grade

•

Hard

•
CCSS
HSA-REI.B.4B, 6.EE.B.7, HSF-IF.C.7A

+3

Standards-aligned

Created by

Elizabeth Ostrowski

Used 7+ times

FREE Resource

10 Slides • 24 Questions

1

16.2 Finding Solutions to Quadratics
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2

Vocabulary

3

Zeros

Zeros, Roots, Solutions, X-Intercepts

The values of X when Y equals zero. The place where the ball hits the ground.

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4

Hotspot

Mark the zeros.

5

The Quadratic Formula

A formula to find the zeros of the parabola.

There will usually be two answers and occasionally one. If you wind up with a negative in the square root, there are no zeros.

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6

Labeling

Drag labels to their correct position on the image
b
ac
a

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8

​Example 1:

9

Multiple Choice

Step 1: Factor x2+10x+21=0x^2+10x+21=0

1

(x+3)(x+7)\left(x+3\right)\left(x+7\right)

2

(x+1)(x+21)\left(x+1\right)\left(x+21\right)

3

(x−7)(x−3)\left(x-7\right)\left(x-3\right)

4

(x−21)(x−1)\left(x-21\right)\left(x-1\right)

10

Dropdown

Step 2: Set both factors equal to zero: (x+3)=\left(x+3\right)= ​

11

Dropdown

Step 2: Set both factors equal to zero: (x+7)=\left(x+7\right)= ​

12

Multiple Choice

Step 3: Solve both factors for x: x+3=0x+3=0 ​​

What should you do to isolate x?

1

Add 0 to both sides

2

Subtract 0 from both sides

3

Add 3 to both sides

4

Subtract 3 from both sides

13

Drag and Drop

x+3=0 and x+7=0x+3=0\ and\ x+7=0

x=​
and x=​
Drag these tiles and drop them in the correct blank above

14

Multiple Choice

x2+x−20=0x^2+x-20=0 is factored into (x+5)(x−4)=0\left(x+5\right)\left(x-4\right)=0 , what are the zeros?

1

x=5 and −4x=5\ and\ -4

2

x=−5 and −4x=-5\ and\ -4

3

x=5 and 4x=5\ and\ 4

4

x=−5 and 4x=-5\ and\ 4

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16

​Example 2:

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Multiple Choice

Step 1: Factor 3x2−5x−2=03x^2-5x-2=0

1

(3x+1)(x−2)\left(3x+1\right)\left(x-2\right)

2

(3x−1)(x+2)\left(3x-1\right)\left(x+2\right)

3

(x−2)(x+1)\left(x-2\right)\left(x+1\right)

4

(3x+3)(x−1)\left(3x+3\right)\left(x-1\right)

18

Dropdown

Step 2: Set both factors equal to zero: (3x+1)=\left(3x+1\right)= ​

19

Dropdown

Step 2: Set both factors equal to zero: (x−2)=\left(x-2\right)= ​

20

Multiple Choice

Step 3: Solve both factors for x: 3x+1=03x+1=0 ​​

What should you do to isolate x?

1

Add 1 to both sides then divide by 3 to make 13\frac{1}{3}

2

Subtract 1 from both sides then divide by 3 to make −13-\frac{1}{3}

3

Add 1 to both sides and multiply by 3 to make 3.

4

Subtract 1 from both sides and multiply by 3 to make -3.

21

Drag and Drop

3x+1=0 and x−2=03x+1=0\ and\ x-2=0

x=​ ​
and x=​ ​
​
Drag these tiles and drop them in the correct blank above

22

Multiple Choice

2x2−5x−12=02x^2-5x-12=0 is factored into (2x+3)(x−4)=0\left(2x+3\right)\left(x-4\right)=0 , what are the zeros?

1

x=32 and −4x=\frac{3}{2}\ and\ -4

2

x=−32 and −4x=-\frac{3}{2}\ and\ -4

3

x=−32 and 4x=-\frac{3}{2}\ and\ 4

4

x=−23 and 4x=-\frac{2}{3}\ and\ 4

23

Quadratic Formula

24

​Example 3:

25

Drag and Drop

Given 2x2+4x−96=02x^2+4x-96=0 , a=​
, b=​
and c=​
Drag these tiles and drop them in the correct blank above

26

Labeling

Given 2x2+4x−96=02x^2+4x-96=0 has a=2a=2 , b=4b=4 and c=−96c=-96 , place the numbers in the correct place in the formula.

Drag labels to their correct position on the image

27

Multiple Choice

Given 2x2+4x−96=02x^2+4x-96=0 , choose the correct quadratic formula.

1

4±42−4(2)(−96)2(2)\frac{4\pm\sqrt[]{4^2-4\left(2\right)\left(-96\right)}}{2\left(2\right)}

2

−4±42−4(2)(−96)2(2)\frac{-4\pm\sqrt[]{4^2-4\left(2\right)\left(-96\right)}}{2\left(2\right)}

3

−4±4−4(2)(−96)2(2)\frac{-4\pm\sqrt[]{4-4\left(2\right)\left(-96\right)}}{2\left(2\right)}

4

−4±42−4(2)(96)2(2)\frac{-4\pm\sqrt[]{4^2-4\left(2\right)\left(96\right)}}{2\left(2\right)}

28

Dropdown

Given −4±42−4(2)(−96)2(2)\frac{-4\pm\sqrt[]{4^2-4\left(2\right)\left(-96\right)}}{2\left(2\right)} , simplify the function: 424^2 becomes ​
, −4(2)(−96)-4\left(2\right)\left(-96\right) becomes ​
and 2(2)2\left(2\right) becomes ​

29

Multiple Choice

Now −4±42−4(2)(−96)2(2)\frac{-4\pm\sqrt[]{4^2-4\left(2\right)\left(-96\right)}}{2\left(2\right)} has become:

1

4±16+7684\frac{4\pm\sqrt[]{16+768}}{4}

2

4±16−7684\frac{4\pm\sqrt[]{16-768}}{4}

3

−4±16+7684\frac{-4\pm\sqrt[]{16+768}}{4}

4

−4±16−7684\frac{-4\pm\sqrt[]{16-768}}{4}

30

Multiple Choice

Since we have −4±16+7684\frac{-4\pm\sqrt[]{16+768}}{4} and 16+768 creates −4±7844\frac{-4\pm\sqrt[]{784}}{4} , what is 784\sqrt[]{784} ?

1

±27\pm27

2

±28\pm28

3

±29\pm29

4

±30\pm30

31

Multiple Choice

Now that we have −4±284\frac{-4\pm28}{4} , what two factions does this make?

1

−4+284and 4+284\frac{-4+28}{4}and\ \frac{4+28}{4}

2

−4+284and 4−284\frac{-4+28}{4}and\ \frac{4-28}{4}

3

4+284and 4−284\frac{4+28}{4}and\ \frac{4-28}{4}

4

−4+284and −4−284\frac{-4+28}{4}and\ \frac{-4-28}{4}

32

Multiple Choice

Since our two fractions are −4+284and −4−284\frac{-4+28}{4}and\ \frac{-4-28}{4} , what does they equal?

1

−6 and 8-6\ and\ 8

2

6 and −86\ and\ -8

3

6 and 86\ and\ 8

4

−6 and −8-6\ and\ -8

33

Dropdown

Given 12x2−12x−6=012x^2-12x-6=0 , a=​
, b=​ ​
and c=​

34

Match

Match the factor with the solution:

x=−12x=-\frac{1}{2}

x=12x=\frac{1}{2}

x=2x=2

x=−2x=-2

x=−23x=-\frac{2}{3}

(2x+1)=0\left(2x+1\right)=0

(2x−1)=0\left(2x-1\right)=0

(x−2)=0\left(x-2\right)=0

(x+2)=0\left(x+2\right)=0

(3x+2)=0\left(3x+2\right)=0

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16.2 Finding Solutions to Quadratics
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